7:30:54 pmAMC 10 PROBLEM SERIESWEEK 7: CIRCLES7:31:26 pmLast week, we looked at one of the fundamental figures in geometry, namely triangles. This week, we look at the other fundamental figure in geometry, namely the circle. We start by looking at problems involving the areas of circles.7:31:34 pm

7:31:56 pmFirst, let's draw a diagram.7:32:01 pm

7:32:11 pmAs a start, let's think about what area Spot can reach if his rope doesn't bend. What does this area look like?7:33:20 pmIf Spot's rope doesn't bend, then the area that Spot can reach is a circular sector, bounded by AB and AF.7:33:26 pm

7:34:23 pmWhat is the area of this circular sector?7:35:28 pmThis circular sector has radius 2 and contains an angle of 240 degrees, so its area is 240/360 = 2/3 the area of a circle with radius 2.7:35:33 pm

7:35:35 pmOf course, Spot's rope can bend. What additional area does this give us?7:36:31 pmIn addition, we get two circular sectors, one bounded by AB and BC, and one bounded by AF and EF. (It's clear that Spot has just enough rope to reach C and E.)7:36:37 pm

7:36:43 pmEach circular sector has radius 1 and contains an angle of 60 degrees, so its area is 60/360 1= 1/6 the area of a circle with radius 1.7:36:49 pm

7:36:51 pmSo what is the total area that Spot can reach?7:37:55 pm

7:37:56 pmThe answer is (E).7:38:17 pm

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7:38:59 pmWe want to find the area of the lune. Let's start simple. Is there an easy shape that we can relate the lune to?7:39:56 pmWe can relate the lune to the semicircle with diameter 1.7:40:44 pm

7:40:46 pmWhat is the area of this semicircle?7:41:27 pm

7:41:34 pmNow, to find the area of the lune, we must subtract the area of the circular segment.7:41:42 pm

7:41:49 pmHow can we find the area of this circular segment?7:43:28 pmFirst, we can draw the lines joining the center of the large semicircle to the endpoints of the circular segment.7:44:14 pm

7:44:56 pmWe can then find the area of the circular segment by finding the area of the circular sector OAB, and subtracting the area of triangle OAB.7:45:01 pmWhat can we say about triangle OAB?7:45:17 pm

7:46:15 pmSo what is the area of circular sector OAB?7:46:20 pm

7:47:12 pm

7:47:16 pm

7:47:31 pmWhat is the area of triangle OAB?7:47:42 pm

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7:48:43 pmWhat does this simplify to?7:49:36 pm

7:49:39 pmThe answer is (C).7:49:49 pmThe problem is very typical for areas involving circles. If we have an odd shape defined in terms of arcs and/or lines, then we can relate the shape to areas we already know how to compute.7:50:37 pm

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7:51:10 pmAs in the previous problem, we want to relate the area we want to other areas that we already know how to find.7:51:25 pmWith this in mind, what can we add to the diagram that will help?7:52:58 pmSince the area we want involves arc DE, we add line segments OD and OE to the diagram.7:53:04 pm

7:53:10 pm(Note that in addition to adding OD and OE, we have left out other elements in the diagram. If we did not leave out those other elements, then we would get a very cluttered diagram, and it would be hard to see what was going on. Thus, a useful technique in solving geometry problems is to draw extra diagrams that contain only the elements that you are interested in.)7:53:50 pmHow can we find the area of circular sector DOE?7:54:57 pm

7:55:01 pm

7:55:04 pmWhat are OA and OD?7:56:18 pm

7:57:11 pm

7:57:14 pm

7:57:17 pm

7:57:22 pm

7:58:03 pm

7:58:04 pmSo what is the area of circular sector DOE?7:58:09 pm

7:59:32 pmCircular sector DOE has radius 2 and contains 30 degrees, so its area is 30/360 = 1/12 the area of a circle of radius 2.7:59:38 pm

8:00:07 pmHow do we find the area we want?8:01:17 pm

8:01:44 pmWe take the area of circular sector DOE, and subtract the area of quadrilateral ODBE.8:01:48 pmWe can split quadrilateral ODBE into triangles OBD and OBE.8:02:42 pm

8:02:44 pmHow can we find the area of triangle OBD?8:05:01 pmThe easiest way of finding the area of triangle OBD is to consider the side BD as the base. Again, we draw another diagram that makes this idea clearer.8:05:06 pm

8:05:10 pmWhat is the height of triangle OBD with respect to base BD?8:06:00 pmThe height of triangle OBD with respect to base BD is OA = 1.8:06:04 pmWhat is the length of base BD?8:06:28 pm

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8:08:10 pmThe answer is (A).8:09:27 pmWe now look at problems involving circles tangent to lines or other circles. Whenever you have circles that are tangent to other geometrical objects, it is almost always useful to join the centers of the circles to the points of tangency.8:09:31 pm

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8:10:32 pmWhat can we do with the diagram?8:11:22 pmWe can join the centers of the circles to the points of tangency.8:11:29 pm

8:12:38 pmAlthough many features of the diagram are evident, such as angles, it's still a good idea to derive what we want rigorously.8:12:41 pmWe start with what we know, namely the three small circles that are tangent to each other.8:12:44 pmWhat can we say about triangle ABC?8:13:36 pmEach of the small circles has radius 1, so each side of triangle ABC has length 2. Therefore, triangle ABC is equilateral.8:13:40 pmLet r be the radius of the large circle. What is OA in terms of r?8:14:33 pmThe circles with centers O and A are tangent at T, so points O, A, and T are collinear.8:14:36 pmHence, OA = OT - AT = r - 1.8:14:40 pmSimilarly, OB = r - 1 and OC = r - 1. What does this tell us about point O in terms of triangle ABC?8:15:49 pmSince OA = OB = OC = r - 1, O is the circumcenter of triangle ABC.8:15:59 pmBut as triangle ABC is equilateral, O is also the centroid of triangle ABC. Hence, O lies on medians AP, BQ, and CR.8:16:11 pm

8:17:15 pmWhat can we say about triangle AOQ?8:18:22 pmSince O is the center of equilateral triangle ABC, and Q is the midpoint of AC, triangle AOQ is a 30-60-90 triangle. Do we know any sides of triangle AOQ?8:19:02 pmWe know that AQ = 1. So what is OA?8:20:16 pm

8:20:34 pmBut we also know that OA = r - 1. So what is r?8:21:27 pm

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8:22:43 pm

8:22:52 pmWhat can we do with the diagram?8:24:20 pmWe can join the centers of the circles to the points of tangency.8:24:24 pm

8:24:30 pmDo we see anything in the diagram that is notable or interesting?8:25:46 pmWe see that both PT and QU are perpendicular to AC, so they are parallel.8:25:56 pmThis tells us that triangles APT and AQU are similar. Do we know any sides of triangles APT or AQU?8:26:57 pmWe know that PT = 1 and QU = 2.8:27:03 pmWe also know that PQ = 1 + 2 = 3. Does this give us any other lengths in the diagram?8:28:39 pmSince PT = 1 and QU = 2, the sides of triangle AQU are double the corresponding sides of triangle APT. Hence, P is the midpoint of AQ.8:28:45 pmBut PQ = 3, so AP = PQ = 3. How does this help in finding the area of triangle ABC?8:30:41 pmWe can say that the altitude of triangle ABC is AM = AP + PQ + QM = 3 + 3 + 2 = 8.8:30:44 pmTo find the area of triangle ABC, we still need BC. How can we find BC?8:32:15 pmSince M is the midpoint of BC, we can find BC by finding CM. How can we find CM?8:32:33 pmNote that CM is a side of triangle AMC. Is there anything we can compare triangle AMC to?8:33:26 pmWe see that triangles AMC and ATP are similar. So what useful equation can we write down?8:34:39 pm

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8:37:06 pmMidpoint of Chords8:37:12 pmFinally, we look at problems that highlight a certain property of circles: If we drop a perpendicular from the center of a circle to a chord, then what can we say about the diagram?8:37:51 pmIf we drop a perpendicular from the center of a circle to a chord, then the foot of the perpendicular is the midpoint of the chord.8:38:03 pm

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8:38:20 pmWe want to find the diameter of the circle. So what can we do with the diagram?8:39:59 pmWhen we want to find lengths, we can try building right triangles.8:40:03 pmWe get right triangles when we drop the perpendiculars from O to chords AB and CD. Drawing in some radii, such as OB, should be useful: drawing radii to relevant points is often a good idea!8:40:09 pm

8:40:15 pmHow does this help us find the diameter of the circle?8:41:22 pmWe can use Pythagoras on right triangle OMB to find the radius OB.8:41:29 pmWhat is BM equal to?8:41:45 pm

8:41:47 pmWhat is OM equal to? Note that we can't calculate it directly.8:43:07 pm

8:43:13 pm

8:43:26 pmThen NP = CP - CN = 6 - 4 = 2. Hence, OM = NP = 2. So what is OB?8:44:57 pm

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8:46:21 pmWhat can we do with the diagram?8:47:34 pmSince AG is tangent to the circle centered at P, we can draw PG.8:48:16 pm

8:48:21 pmHow can we find the length of chord EF?8:50:02 pmWe can find lengths by building right triangles.8:50:06 pmWe drop the perpendicular from center N to chord EF.8:50:13 pm

8:50:22 pmHow does this help us find the length of chord EF?8:51:47 pmWe know that M is the midpoint of EF, so it suffices to find EM. How can we find EM?8:53:10 pmWe can apply Pythagoras on right triangle EMN to find EM. But to do this, we must find EN and MN. What is EN?8:53:37 pmWe know that EN = 15, the radius of the circle. How can we find MN?8:54:05 pmNote that MN is a side of triangle AMN.8:54:10 pmSince both MN and PG are perpendicular to AG, they are parallel.8:54:15 pmHence, triangles AMN and AGP are similar. So what useful equation can we write down?8:55:26 pm

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8:56:43 pm

8:56:51 pm

8:56:57 pm

8:57:03 pmWe can start by drawing a diagram. However, we must be careful to draw an accurate diagram. For example, the diagram could like either of the following.8:57:18 pm

8:57:25 pmCan we tell which is the right diagram?8:58:12 pm

8:58:15 pm

8:58:26 pmWe can drop the perpendicular from O to chord BC.8:58:31 pm

8:58:38 pmLet s be the side length of equilateral triangle ABC. What lengths can we write in terms of s?8:59:51 pm

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9:03:32 pmSUMMARY9:03:37 pmIn today's class we looked at many problems involving the properties of circles, such as the areas of circles. If we have an area involving arcs of circles that we do not know how to compute directly, then the best strategy is to compare the area to other figures where you know how to compute the area.9:03:41 pmWe also saw that if you have circles that are tangents to lines or other circles, then it is very useful to join the centers to the points of tangency. Like joining the centers to midpoints of chords, constructing these additional lines can help you build right triangles, which will in turn help you find lengths.

7:30:17 pmAMC 10 PROBLEM SERIESWEEK 8: POLYGONS AND THREE-DIMENSIONAL GEOMETRY7:30:29 pmIn today's class, we continue our look at Euclidean geometry by dealing with problems involving polygons and three-dimensional geometry. In both cases, we draw on the concepts that we have seen in the previous classes.7:30:33 pmPOLYGONS7:30:40 pmIn most problems involving polygons, we depend on the same techniques that we have used before in other geometry problems, such as angle chasing and using right triangles and similar triangles.7:30:49 pm

7:31:01 pmFirst, let's draw a diagram.7:31:05 pm

7:31:09 pmWhat can we say about the diagram?7:32:59 pm

7:33:03 pm

7:33:06 pmAre there any other angles we can express in terms of x?7:34:19 pm

7:34:49 pm

7:34:51 pmIs there anything we can say about the diagram now?7:35:47 pm

7:35:50 pmSince CD = 6, CM = 6.7:35:55 pm

7:36:11 pmNow what can we say about the diagram?7:38:18 pm

7:39:55 pm

7:40:07 pmSolving for x, we find x = 75. The answer is (E).7:40:13 pm

7:41:10 pmFirst, let's draw a diagram. What are the parallel sides in trapezoid ABCD?7:41:45 pmSince both AB and CD are perpendicular to AD, they are parallel.7:41:50 pm

7:41:53 pmWe want to find the product AB * CD. We don't see any way of calculating AB or CD directly, so what can we do?7:43:30 pmWe can use variables. Let x = AB and y = CD.7:43:34 pm

7:43:39 pmWhat else can we express in terms of x and y?7:44:54 pmWe are given that BC = AB + CD, so BC = x + y.7:44:59 pm

7:45:50 pmWhat piece of information haven't we used yet?7:46:09 pmWe haven't used the fact that AD = 7. How can we use this?7:46:29 pmWe can drop a perpendicular from B to CD.7:46:35 pm

7:47:23 pmWhat can we say about the diagram?7:47:43 pmSince ABED is a rectangle, BE = AD = 7 and DE = AB = x. Then CE = CD - DE = y - x.7:47:54 pm

7:48:36 pmNow what can we say about the diagram?7:48:59 pm

7:50:17 pmThis simplifies to 4xy = 49. So what is AB * CD?7:50:28 pmWe have that AB * CD = xy = 49/4 = 12.25. The answer is (B).7:50:33 pm

7:51:35 pmLet's start by drawing a diagram:7:51:40 pm

7:52:36 pmHow can we find the areas of trapezoids ABEF and FECD?7:53:46 pmWe can use the formula for the area of a trapezoid. First, let's label the bases.7:54:25 pm

7:54:30 pmSince E and F are the midpoints of BC and AD, respectively, EF = (x + y)/2.7:54:37 pm

7:55:37 pmWhat can we say about the heights of trapezoids ABEF and FECD?7:56:15 pmThey are equal, so we can let them be h.7:56:20 pm

7:56:56 pmWhat is the area of trapezoid ABEF?7:58:21 pm

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7:59:04 pmMultiplying both sides by 3x + y, we get x + 3y = 2(3x + y) = 6x + 2y. What does this simplify to?8:00:26 pmThis simplifies to y = 5x. We see that AB/DC = y/x = 5. The answer is (C).8:00:35 pm

8:01:55 pmLet's start by drawing a diagram:8:02:00 pm

8:02:02 pmHow can we start?8:03:07 pmWe can start by naming the side length of the regular octagon. Let s be side length of the regular octagon.8:03:11 pmWhat can we add to the diagram that will help?8:03:36 pmWe can add diagonals AF, BE, CH, and DG.8:03:42 pm

8:03:44 pmBy drawing these diagonals, we have created right triangles which we can use to find lengths.8:03:50 pmFor example, what is AP?8:05:59 pm

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8:07:15 pm

8:07:17 pmWhat is the area of rectangle ABQP?8:09:00 pm

8:09:05 pm

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8:11:40 pmBut what else do we know?8:12:05 pm

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8:13:20 pmWhat is AF?8:14:25 pm

8:14:51 pmSo what is the area of rectangle ABEF?8:15:50 pm

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8:16:23 pmThe answer is (D).8:16:59 pmHere is another way to solve the problem.8:17:02 pmWe draw diagonals AE and BF.8:17:07 pm

8:17:21 pmNote that the area of rectangle ABEF is four times the area of triangle OEF.8:17:25 pmBut how does the area of triangle OEF compare to the area of regular octagon ABCDEFGH?8:18:42 pmSince O is the center of the octagon, the area of triangle OEF is 1/8th the area of regular octagon ABCDEFGH.8:18:46 pmTherefore, the area of rectangle ABEF is half the area of regular octagon ABCDEFGH, so it has area 1/2.8:19:12 pmThat's another way as well.8:19:17 pm

8:19:29 pmFirst, let's draw a diagram. What can we say about hexagon ABCDEF that will help us draw it?8:20:28 pmWe are given that hexagon ABCDEF is equiangular, so every angle is 120 degrees.8:20:32 pm

8:20:38 pmWe are given that the area of triangle ACE is 70% the area of hexagon ABCDEF. To turn this into a useful equation, we must express both areas in terms of r. How can we do that?8:21:43 pmIs there anything we can do with the figure that will help us find the area of hexagon ABCDEF?8:23:01 pmWe can extend the sides of the hexagon to form an equilateral triangle.8:23:06 pm

8:23:15 pmWhat is the area of equilateral triangle PQR?8:25:07 pm

8:25:30 pmHow can we get the area of hexagon ABCDEF?8:27:07 pmWe can get the area of hexagon ABCDEF by subtracting the areas of equilateral triangles PAB, QCD, and REF.8:27:17 pm

8:28:09 pmHow can we find the area of triangle ACE?8:28:16 pm

8:29:36 pmWe can take the area of equilateral triangle PQR, and subtract the areas of triangles PAC, QCE, and REA.8:29:42 pmHow can we find the area of triangle PAC?8:32:13 pmWe can use the formula from the triangle class, comparing the areas of two triangles.8:32:20 pm

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8:34:48 pmWe can now use the fact that the area of triangle ACE is 70% the area of hexagon ABCDEF.8:35:20 pm

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8:36:17 pmWhat does this simplify to?8:37:43 pm

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8:38:16 pmWhat are the roots of this equation?8:40:16 pm

8:40:24 pmNotice that both of these numbers are positive and therefore acceptable.8:40:29 pm

8:40:36 pmWe could have computed the sum of the roots using Vieta's formulas, but how do we know that both roots are possible values? What if one of the roots is negative? By computing both roots and seeing that they are positive, we see that they are viable.8:40:50 pmTHREE-DIMENSIONAL GEOMETRY8:40:55 pmWe start our exploration of three-dimensional geometry by working with common solids, such as pyramids, cones, and spheres.8:40:59 pm

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8:41:23 pmThe circular sector has radius 10. What length does this correspond to in the cone?8:42:49 pmThis corresponds to the slant height of the cone.8:42:53 pmWhat is the circumference of the circular sector?8:44:25 pm

8:44:27 pmThe circular sector contains an angle of 252 degrees, which is 252/360 = 7/10 of the full circle.8:44:31 pm

8:45:09 pmWhat does this correspond to in the cone?8:46:23 pmThis corresponds to the circumference of the base of the cone. So what is the radius of the base of the cone?8:47:24 pm

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8:47:37 pmWe are looking for a cone with slant height 10 and base radius 7, so the correct answer is (C).8:47:46 pm

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8:48:57 pmFirst, let's work with the outer cube. What is the side length of the outer cube?8:50:22 pmThe outer cube has surface area 24 (in square meters), so each face has area 24/6 = 4. Hence, the side length of the outer cube is 2.8:50:50 pmNext, we work with the sphere. What is the radius of the sphere?8:51:48 pmThe sphere is inscribed in a cube with side length 2. Then the diameter of the sphere is 2, so its radius is 1.8:51:51 pmNow, we need the side length of the inner cube. Let x be the side length of the inner cube. How can we determine x?8:53:21 pmPerhaps the easiest way is to consider a diagonal of the inner cube going from one vertex to an opposite vertex.8:53:26 pm

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8:53:40 pmThis diagonal is also equal to a diameter of the sphere, which is 2.8:54:11 pm

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8:55:06 pmLet the radius of the sphere of vanilla ice cream be r. What is its volume?8:56:18 pm

8:56:22 pmWe know the radius of the cone is the same as the radius of the ice cream, namely r. Let h be height of the cone. Then what is the volume of the cone?8:57:33 pm

8:57:40 pmWe are told that when the ice cream melts, it fills the cone exactly. We are also told that the melted ice cream takes up 75% the volume of the frozen ice cream. So what equation can we write down?8:59:28 pm

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9:00:22 pmSo what is the ratio h/r?9:00:39 pmWe see that h/r = 3. The answer is (B).9:00:45 pmSUMMARY9:00:52 pmIn today's lecture, we looked at techniques to solve problems involving polygons and three-dimensional geometry. However, none of these techniques were really new.9:00:59 pmFor example, when dealing with polygons, one useful approach is to break them up into triangles, and then apply what we know about triangle geometry. In three-dimensional geometry, we can take cross-sections, and use the usual two-dimensional geometry techniques, such as building right triangles and looking for similar triangles.

7:30:39 pmAMC 10 PROBLEM SERIESWEEK 9: COUNTING7:30:46 pmIn today's class, we will cover techniques for solving problems in counting. Counting problems come in a wide variety of forms, and accordingly there are a wide variety of techniques for solving them. We will try to cover as many of these techniques as we can.7:30:50 pmPRODUCT PRINCIPLE7:31:16 pmOne of the simplest principles in counting is the product principle. Suppose I own three shirts, four pairs of pants, and two pairs of shoes. Then how many different outfits can I wear?7:31:30 pmI just multiply the number of choices for each piece of clothing, giving me 3 * 4 * 2 = 24 different outfits.7:31:46 pmMore generally, if I want to choose a number of objects, then the number of ways is simply the product of the numbers of ways of choosing each individual object.7:31:52 pm

7:32:49 pmWhat do we need to compute?7:34:07 pmWe need to compute the number of license plates under the old scheme, and the number of license plates under the new scheme.7:34:38 pmHow many license plates are there under the old scheme?7:35:03 pm

7:35:04 pmHow many license plates are there under the new scheme?7:36:14 pm

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7:36:21 pmThe answer is (C).7:36:27 pm

7:37:51 pmThe problem itself tells us how to apply the product principle. Each tourist must choose one of the two guides.7:37:55 pmSo the first tourist chooses one of the two guides, the second tourist chooses one of the two guides, and so on. How many groupings does this give us?7:38:44 pm

7:39:03 pmThe problem also tells us that in each grouping, each guide must take at least one tourist. So we must subtract the number of groupings in which a guide has no tourists. How many such groupings are there?7:39:31 pmIf a guide has no tourists, then the other guide gets all the tourists. There are two ways this can happen, one for each guide.7:39:34 pmHence, there are 64 - 2 = 62 possible groupings. The answer is (D).7:39:38 pmWhen we obtain a number in a counting problem, it is common that because of some condition in the problem, we must adjust the answer somehow. In any counting problem, make sure you read the problem carefully for any particular conditions.7:40:50 pm

7:41:07 pmHow can we start?7:42:21 pmLet a, b, and c be the three digits. We want these digits to satisfy (a + c)/2 = b, or a + c = 2b.7:42:27 pmEquivalently, the digits a, b, and c form an arithmetic sequence. How can we count the number of such arithmetic sequences?7:44:35 pmThere are many ways to approach this problem. For example, we can ask how many arithmetic sequences have a first term of 0, a first term of 1, and so on.7:44:41 pmWe can also ask how many arithmetic sequences have a certain common difference. For example, we can ask how many arithmetic sequences have a common difference of 0, a common difference of 1, and so on. (And don't forget negative common differences.)7:45:43 pmBoth approaches will work, but they involve dividing into cases. Let's see if we can find an approach that avoids casework.7:45:48 pmTo get a feel for the problem, let's look at a certain example. How many such three-digit numbers start with the digit 3?7:46:21 pmThe three-digit numbers that start with the digit 3 are 321, 333, 345, 357, and 369. Do you see anything interesting about these numbers?7:47:28 pmWhat is interesting to me about these numbers are the last digits, namely 1, 3, 5, 7, and 9.7:47:35 pmThese digits are all odd. Is there something we can say more generally?7:49:11 pmIn every three-digit number that we want to count, the first and last digits are either both even or both odd. (This follows from the equation a + c = 2b.)7:49:17 pmSo let's look at the even case and odd case. (So we are using casework, but it's only two cases, so it's not that bad.)7:49:23 pmHow many possible first digits are even?7:50:20 pmThe possible first digits that are even are 2, 4, 6, and 8 (but not 0), for a total of 4. How many possible last digits are even?7:51:15 pmThe possible last digits that are even are 0, 2, 4, 6, and 8, for a total of 5.7:51:20 pmSo how many ways can we choose the first and last digits to be even?7:51:45 pmWe can choose the first and last digits to be even in 4 * 5 = 20 different ways.7:52:28 pmNotice that when both first and last digits are even, for any choice of the first and last digits the middle digit is uniquely determined.7:52:32 pmNow we look at the odd case. How many possible first digits are odd?7:52:44 pmThe possible first digits that are odd are 1, 3, 5, 7, and 9, for a total of 5.7:52:47 pmThe possible last digits that are odd are the same, for another 5. So how many ways can we choose the first and last digits to be odd?7:53:01 pmWe can choose the first and last digits to be odd in 5 * 5 = 25 different ways.7:53:08 pmSimilar to the even case notice that when both first and last digits are odd, for any choice of the first and last digits the middle digit is uniquely determined.7:53:11 pmSo how many such three-digit numbers are there?7:53:23 pmThe total number of such three-digit numbers is 20 + 25 = 45. The answer is (E).7:53:27 pmNot every counting problem can be solved by using the product principle, but if you can use it, then it is usually the easiest way to solve the problem.7:53:43 pmCOMBINATIONS7:53:56 pmIn a combination, we are selecting objects from a set, where the order of the objects does not matter. For example, how many ways can we select three different letters of the alphabet? The order of the letters does not matter, so selecting A, B, and C is the same as selecting B, C, and A.7:55:56 pm

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7:56:25 pmThis symbol is known as a binomial coefficient, and it is read "n choose k." In counting, many answers depend in some form on a combination.7:56:38 pm

7:56:49 pmHow can we specify a line segment, where both endpoints are vertices of the cube?7:57:31 pmWe specify such a line segment by choosing two vertices of the cube. How many ways are there to choose two vertices of the cube?7:57:55 pm

7:58:03 pm

7:58:37 pmWhat do we want to compute in this problem?8:00:39 pmWe want to compute the number of ways of arranging three Xs and two Os in a row. How can we do this?8:01:12 pmIt's not hard to simply list by hand all the ways of arranging three Xs and two Os in a row, but using combinations makes things even easier. How can we use combinations?8:01:56 pmLet's think about how can specify an arrangement of three Xs and two Os in a row. We have five tiles to order, so how can we specify such an arrangement?8:02:48 pmWe can specify such an arrangement by placing the three Xs first (or we can go with the Os if you want -- same idea.)8:02:54 pm

8:02:59 pmTherefore, the probability that the arrangement reads XOXOX is 1/10. The answer is (B).8:03:10 pmSTARS AND BARS8:03:32 pmWe now present a method for counting certain kinds of distributions, which we call stars and bars. (It goes by other names too, like balls and urns.)8:03:40 pmTo illustrate the method, we look at a specific example.8:03:46 pm

8:06:26 pmFor example, Pat can buy two glazed, one chocolate, and one powdered. It's not hard to list all the possible ways, but stars and bars make this problem a snap.8:06:32 pmTo apply stars and bars, we represent each donut with a star, and we divide the groups of donuts (by type) with bars. For example, if Pat buys two glazed, one chocolate, and one powdered, then we represent this with the following row of stars and bars:8:06:47 pm* * | * | *8:06:54 pmThe first, second, and third groups represent glazed, chocolate, and powdered, respectively. What does the following row represent?8:07:00 pm* * * | | *8:08:24 pmThis row represents three glazed and one powdered. There are no stars between the two dividers, so there are no chocolate donuts.8:08:33 pmThus, every combination of four donuts can be represented by a row of four stars and two bars. How many ways can we arrange four stars and two bars?8:09:46 pm

8:09:52 pmTherefore, Pat can make 15 different selections. The answer is (D).8:10:01 pmCOMPLEMENTARY COUNTING8:10:45 pmIn some counting problems, it is easier to count the objects that we don't want, rather that the objects that we do want. This is technique is called complementary counting.8:11:13 pm

8:11:46 pmWhat is the difficult part about this problem?8:12:37 pmThe difficult part about this problem is that there are many ways that a four-digit number that can have at least one digit that is a 2 or a 3. For example, the number can be 7201, or 3587, or 2322.8:13:23 pmThe number can have a 2, or a 3, or a 2 and a 3, leading to many possible combinations. It is difficult to know where to start. So what can we try?8:14:21 pmWe can try looking at numbers that do not satisfy the given condition.8:14:28 pmIf the number does not satisfy the given condition, then the number does not contain at least one digit that is a 2 or a 3. In other words, none of the digits are 2 or 3.8:14:38 pmAlready this seems like an easier condition to deal with. How can we count the number of four-digit number where none of the digits are 2 or 3?8:16:25 pmWe look at each digit individually. What are the possible first digits?8:17:33 pmThe possible first digits are 1, 4, 5, 6, 7, 8, and 9, for a total of 7. What are the possible second digits?8:18:45 pmThe possible second digits are 0, 1, 4, 5, 6, 7, 8, and 9, for a total of 8.8:19:15 pmThe possible third and fourth digits are the same as the possible second digits. So how many four-digit numbers do not contain a 2 or 3?8:19:40 pmThe number of four-digit numbers that do not contain a 2 or 3 is 7 * 8 * 8 * 8 = 3584. What do we want to do with this number?8:20:32 pmWe want to subtract it from the total number of four-digit numbers.8:20:43 pmUsing the same method of counting digit by digit, we find there are 9 * 10 * 10 * 10 = 9000 four-digit numbers.8:20:52 pmSubtracting, we get 9000 - 3584 = 5416 numbers. The answer is (E).8:20:59 pmCASEWORK8:21:14 pmAs much as we've tried to avoid using casework, in many counting problems, it is inevitable. So let's work through a counting problem using casework.8:21:29 pm

8:22:54 pmHow can we get started?8:24:50 pmAs is customary in society, we seat the ladies first. How many ways can the first woman be seated?8:25:49 pmThe first woman can be seated in ten ways, in each of the ten chairs. Where can the remaining four women sit?8:26:31 pmThe men and women must alternate, so there are only four chairs that the remaining four women can sit in. How many ways are there to seat the remaining four women?8:27:59 pmThere are 4! = 24 ways to seat the remaining four women.8:28:02 pmWe label the five women as follows.8:29:11 pm

8:29:17 pmNext, we look at the men. Assume that the married couples are W1 and M1, W2 and M2, and so on.8:29:22 pmLet's look at man M1. Where can man M1 sit?8:30:26 pmNo one can sit next to or directly across from his or her spouse, so man M1 can only sit between W2 and W3, or W4 and W5.8:30:59 pmLet's consider the case where M1 sits between W2 and W3.8:31:04 pm

8:31:10 pmIs there anything we can say that would help us fill in the diagram?8:32:25 pmMan M4 can only sit between women W1 and W5.8:33:01 pm

8:33:04 pmNow what can we say about the diagram?8:33:53 pmMan M2 can only sit between women W3 and W4.8:33:57 pm

8:34:03 pmThen man M5 must sit between women W1 and W2, and then man M3 must sit between women W4 and W5.8:34:45 pm

8:35:15 pmSo after man M1 sits between women W2 and W3, there is only one way to seat the rest of the men.8:35:19 pmWhat if man M1 sits between women W4 and W5? How many ways are there to seat the rest of the men then?8:35:24 pm

8:36:10 pmThe case is symmetric with the first case, so again there is only one way to seat the rest of the men, which we can derive as follows.8:36:22 pm

8:36:29 pmLet's recap.8:36:35 pmThere are 10 ways to seat the first women, then 24 ways to seat the remaining women.8:36:39 pmThen there are 2 ways to seat the first man, then only one way to seat the remaining men. So how many seating arrangements are there?8:38:09 pmThere are 10 * 24 * 2 = 480 different seating arrangements. The answer is (C).8:38:13 pmWhen dividing into cases, make sure that your cases are exhaustive (in other words, your cases cover all possibilities), and that you work all your cases through to the end. Sometimes, it may be necessary to divide your cases into sub-cases. The key is to be diligent and thorough.8:38:19 pmTHINK ABOUT IT!8:38:32 pmThe last technique we want to discuss is called "think about it." When solving a counting problem (or any problem in mathematics), it is easy to become conditioned to look for the right formula or technique that seems to fit the situation. However, in some counting problems, the best way to solve the problem is to put your pencil down and simply think about what it is really asking for.8:39:08 pm

8:40:11 pmHow can we solve this problem?8:41:36 pmWe can think about how the tournament proceeds, and the number of players left in each round.8:41:40 pmWe are told that the top 28 players get a bye, meaning they automatically get to the second round. How many games do the 72 remaining players play?8:42:32 pmThe remaining players play 72/2 = 36 games. So how many players advanced to the second round?8:42:37 pmA player is eliminated in each of the 36 games, so 36 + 28 = 64 players advance to the second round.8:43:41 pmWe can continue to analyze the tournament round by round, counting the number of games played, and the number of players who advance. But there is a much faster way.8:43:46 pmThe tournament goes from 100 players down to 1 player, so we need to eliminate 99 players. How does a player get eliminated?8:44:44 pmExactly one player gets eliminated in each game. So how many games must there be?8:45:12 pmThere must be 99 games. Of the answer choices, only (E) is true. That's it.8:45:40 pmThe fact that the top 28 players get a bye is completely irrelevant. This condition impacts the number of rounds in the tournament, but it has no effect on the number of matches that must be played.8:45:47 pm

8:46:19 pmHow can we start?8:46:56 pmWe can start by drawing a diagram.8:47:03 pm

8:48:14 pmWe get... a messy diagram. Trying to count all the triangles based on the diagram is going to be difficult, if not impossible. So what can we do?8:48:41 pmOuch, no.8:49:02 pmSmart but they won't let you do it during the AMC8:49:37 pmThat's a lot of letters. We're going for simple.8:50:03 pmWe can now look at one individual triangle, and see if we can determine anything interesting.8:50:10 pm

8:51:08 pmSince we are trying to count the number of triangles, we should figure out how such a triangle is created in the first place. How does this triangle get created?8:51:17 pmThis triangle is created by three chords.8:51:27 pm

8:52:33 pmIn turn, the three chords are specified by six points on the circle. So how can we count the number of triangles?8:54:05 pmEvery triangle is created by choosing six points on the circle. Furthermore, it is not hard to see that there is only one triangle that can be created from these six points. In order to form a triangle with the vertices in the interior of the circle, each of the 6 points must be connected to the one opposite.8:54:13 pm

8:54:34 pmSUMMARY8:54:37 pmToday, we saw many concepts for solving problems in counting. You may be wondering, when we face a problem in counting, how do we know which concept is the right one to apply?8:54:43 pmDo not start, for example, by trying to guess the right binomial coefficient that seems to fit. The right way to start solving a counting problem is to consider what it is you are trying to count. Is there a simple way of describing the objects you want to count? Is there a simple process that generates the objects? Only after you find the right approach should you start working with the actual numbers.8:54:56 pmThe most important thing is to keep your approach as simple as possible. Start by looking for a way to use the product principle or combinations. If these do not work, you can always try complementary counting and casework. And last but not least... just think about it.Valentin Vornicu7:30:58 pmAMC 10 PROBLEM SERIESWEEK 10: PROBABILITYValentin Vornicu7:31:05 pmIn today's lecture, we look at various types of probability problems and the techniques for solving them. We start with the most common definition of probability.Valentin Vornicu7:31:10 pmPROBABILITY BY COUNTINGValentin Vornicu7:32:05 pmIn probability, we typically work with "events." For example, flipping a coin and getting three heads in a row can be an event.Valentin Vornicu7:32:10 pmIn situations where all outcomes are regarded as equally likely, the probability of an event is the number of ways that event can occur divided by the total number of possible outcomes. Thus, we use counting when applying this definition.Valentin Vornicu7:32:16 pm

Valentin Vornicu7:32:37 pmWhat do we need to compute in this problem?Valentin Vornicu7:33:23 pmWe need to compute the number of integers between 1 and 100 that are divisible by 2 but not divisible by 3.Valentin Vornicu7:33:26 pmHow many of these numbers are divisible by 2?Valentin Vornicu7:33:45 pmWe see that 100/2 = 50 of these numbers are divisible by 2. These numbers are {2, 4, 6, ..., 100}.Valentin Vornicu7:33:57 pmHow many of these numbers are not divisible by 3?Valentin Vornicu7:34:16 pmIt is not so easy to determine how many of the numbers {2, 4, 6, ..., 100} are not divisible by 3. Instead of trying to find how many are not divisible by 3, what else can we ask?Valentin Vornicu7:35:25 pmWe can ask how many of the numbers {2, 4, 6, ..., 100} are divisible by 3.Valentin Vornicu7:35:32 pmA number is divisible by both 2 and 3 if and only if it is divisible by 6. How many numbers between 1 and 100 are divisible by 6?Valentin Vornicu7:35:47 pmWhen we divide 100 by 6, we get 16 with a reminder of 4, so there are 16 multiples of 6 between 1 and 100.Valentin Vornicu7:36:14 pmSo how many of the numbers {2, 4, 6, ..., 100} are not divisible by 3?Valentin Vornicu7:36:40 pmSince 16 of these numbers are divisible by 3, 50 - 16 = 34 of these numbers are not divisible by 3. So what is the answer?Valentin Vornicu7:37:07 pmThe probability that a number from 1 to 100 is divisible by 2 and not divisible by 3 is 34/100 = 17/50. The answer is (C).Valentin Vornicu7:37:14 pm

Valentin Vornicu7:38:30 pmSo we'll need to count the numerator and denominator. But before we do that, we have to decide a couple things: Are the bills distinguishable? Does order matter? No matter what we choose, the probability will be the same (it better be). We would get different counts that have the same ratio.Valentin Vornicu7:39:02 pmSo the question is: what choices do we make to make counting easy? Let's try distinguishable, order does matter.Valentin Vornicu7:39:10 pmWhat is the denominator of the probability?Valentin Vornicu7:40:06 pmThere are 8 possibilities for the first bill, and 7 possibilities for the second bill, for a total of 56.Valentin Vornicu7:40:26 pmWhat is the numerator of the probability?Valentin Vornicu7:41:05 pmThe numerator is the number of ways of choosing two bills so that the sum of the bills is $20 or more. Which combinations of bill will work?Valentin Vornicu7:42:55 pmWe need at least one twenty or two tens. Sounds like casework!Valentin Vornicu7:43:27 pmLet's start with the two tens. If the two tens are distinguishable and order matters, how many counts is that?Valentin Vornicu7:44:33 pmWe can draw 10A and then 10B or vice versa. So that's two cases.Valentin Vornicu7:44:37 pmTwo twenties also has 2 possibilities.Valentin Vornicu7:44:46 pmWhat about if we have two different bills? Let's say one and twenty. How many possibilities is that?Valentin Vornicu7:45:17 pmWe can draw a twenty then a smaller bill, or vice versa. Also, we have 2 possibilities for which 20 we draw and 6 possibilities for which smaller denomination bill we draw. How many possibilities is this altogether?Valentin Vornicu7:47:06 pmThere are 2*2*6=24 possibilities for drawing bills of different denomination with total value over 20 dollars. Altogether there are 24 + 4 = 28 successful outcomes. Since there were 56 outcomes in total, the probability is 28/56 = 1/2.Valentin Vornicu7:47:48 pm

Valentin Vornicu7:48:30 pmFirst, we look at the probability p. What is the numerator of p?Valentin Vornicu7:49:32 pmThe numerator of p is the number of ways of choosing four slips so that they all have the same number.Valentin Vornicu7:49:35 pmThere are only 10 ways to do this, one for each number from 1 to 10.Valentin Vornicu7:49:38 pmWhat is the denominator of p?Valentin Vornicu7:50:16 pmThe denominator of p is the number of ways of choosing four slips. But do we need to compute this number?Valentin Vornicu7:50:58 pmThe denominators of p and q are the same, so when we compute q/p, they cancel. Hence, there is no need to compute the denominator of p. Let's just call this large number N.Valentin Vornicu7:51:01 pmWhat is the numerator of q?Valentin Vornicu7:52:29 pmThe numerator of q is the number of ways of choosing four slips of the form a, a, b, b, where a and b are different numbers. How can we start counting this number?Valentin Vornicu7:53:22 pmFirst, we can determine the number of ways of choosing the numbers a and b. How many ways can we choose the numbers a and b?Valentin Vornicu7:54:25 pmThe number of ways of choosing the numbers a and b is 10 choose 2 = 45.Valentin Vornicu7:54:42 pmNow that we have determined a and b, we must determine the number of ways of choosing two slips with the number a, and two slips with the number b.Valentin Vornicu7:54:45 pmHow many ways can we choose two slips with the number a?Valentin Vornicu7:56:06 pmSince there are four slips with the number a, we can choose two slips with the number a in 4 choose 2 = 6 ways.Valentin Vornicu7:56:13 pmSimilarly, we can choose two slips with the number b in 4 choose 2 = 6 ways.Valentin Vornicu7:56:17 pmThe numerator of q is 45 * 6 * 6 = 1620. So what is q/p?Valentin Vornicu7:57:42 pmWe see that q/p = (1620/N)/(10/N) = 162. The answer is (A).Valentin Vornicu7:58:22 pmALGEBRAIC PROBABILITYValentin Vornicu7:58:25 pmIn some probability problems, we must use the algebraic properties of probability. For example, if we toss a coin and roll a die, then what is the probability that we get heads and we roll a number that is less than or equal to 2?Valentin Vornicu7:59:20 pmThe probability that we get heads is 1/2, and the probability that we roll a number that is less than or equal to 2 is 2/6 = 1/3, so the probability that both occur is 1/2 * 1/3 = 1/6.Valentin Vornicu7:59:24 pmIn this example, we are using theindependenceproperty of probability: if two events are independent, then the probability that both occur is the product of their probabilities.Valentin Vornicu7:59:30 pm

Valentin Vornicu8:00:01 pmWhat do we want to do in this problem?Valentin Vornicu8:01:20 pmWe want to find the value of p.Valentin Vornicu8:01:25 pmWe can choose any number from 1 to 100.Valentin Vornicu8:01:31 pmConsider the probability of choosing 1, the probability of choosing 2, and so on, up to the probability of choosing 100. What can we say about all these probabilities that will help us find p?Valentin Vornicu8:02:32 pmHint: they are NOT all equal.Valentin Vornicu8:03:36 pmAll of these probabilities must add up to 1, because they cover all the possible values.Valentin Vornicu8:03:54 pmThe probabilities for each of the numbers from 1 to 50 is p, and the probabilities from each of the numbers from 51 to 100 is 3p. So what is the sum of these probabilities?Valentin Vornicu8:04:29 pmThe sum of these probabilities is 50 * p + 50 * 3p = 50p + 150p = 200p. So what is p?Valentin Vornicu8:05:32 pmThe sum of these probabilities is 1, so 200p = 1, which means p = 1/200.Valentin Vornicu8:05:58 pmWe want to compute the probability of choosing a perfect square. How can we compute the probability of choosing a perfect square?Valentin Vornicu8:07:16 pmUsually, we might compute this probability by finding the number of perfect squares from 1 to 100, then dividing by 100. However, this won't work here.Valentin Vornicu8:08:39 pmBecause not every number from 1 to 100 has the same probability of being chosen. The probability of choosing a number from 1 to 50 is p, and the probability of choosing a number from 51 to 100 is 3p. So what must we do?Valentin Vornicu8:09:33 pmWe must find the number of perfect squares from 1 to 50, and the number of perfect squares from 51 to 100.Valentin Vornicu8:09:36 pmHow many perfect squares are between 1 and 50?Valentin Vornicu8:09:48 pm

Valentin Vornicu8:09:49 pmHow many perfect squares are between 51 and 100?Valentin Vornicu8:10:10 pm

Valentin Vornicu8:10:40 pmSo what is the probability of choosing a perfect square?Valentin Vornicu8:11:37 pm

Valentin Vornicu8:12:13 pmThe answer is (C).Valentin Vornicu8:12:17 pm

Valentin Vornicu8:13:33 pmIn how many ways can exactly one voter approve the mayor's work?Valentin Vornicu8:15:15 pmThere are three ways exactly one voter approve the mayor's work, one for each voter.Valentin Vornicu8:15:20 pmSuppose the first voter approves the mayor's work, and the other two do not. What is the probability of this occurring?Valentin Vornicu8:16:01 pmIf there is a 70% chance that a voter approves the mayor's work, then there is a 30% chance that a voter does not approve the mayor's work.Valentin Vornicu8:16:04 pmHence, the probability that the first voter approves the mayor's work and the other two do not is 0.7 * 0.3 * 0.3 = 0.063.Valentin Vornicu8:16:09 pmThe other two scenarios have the same probability of occurring. So what is the answer?Valentin Vornicu8:17:41 pmThe probability that exactly one voter approve the mayor's work is 3 * 0.063 = 0.189. The answer is (B).Valentin Vornicu8:17:54 pmTREE ANALYSISValentin Vornicu8:17:58 pmMany probability problems involve a process, such as a coin being flipped over and over again, or balls being drawn from an urn. One systematic way of dealing with such processes is constructing a tree, where the branches represent different possible outcomes.Valentin Vornicu8:18:09 pm

Valentin Vornicu8:18:51 pmJacob starts with a 6. What are the possible values of the second term?Valentin Vornicu8:20:44 pmThe possible values of the second term are 2 * 6 - 1 = 11 and 6/2 - 1 = 2.Valentin Vornicu8:20:47 pmAccordingly, we draw a tree with a 6 at the top, branching to 11 and 2.Valentin Vornicu8:20:52 pm

Valentin Vornicu8:22:03 pmIf 11 is the second term, then what are the possible values of the third term?Valentin Vornicu8:22:57 pmThe possible values of the third term are 2 * 11 - 1 = 21 and 11/2 - 1 = 9/2. We add these values to the tree.Valentin Vornicu8:23:01 pm

Valentin Vornicu8:23:04 pmIf 2 is the second term, then what are the possible values of the third term?Valentin Vornicu8:23:23 pmThe possible values of the third term are 2 * 2 - 1 = 3 and 2/2 - 1 = 0.Valentin Vornicu8:23:28 pm

Valentin Vornicu8:24:18 pmThen for each possible third term, we compute the possible fourth terms.Valentin Vornicu8:24:23 pm

Valentin Vornicu8:24:28 pmSo what is the probability that Jacob's fourth term is an integer?Valentin Vornicu8:24:50 pmOf the eight possible fourth terms, five are integers, so the probability is 5/8. The answer is (D).Valentin Vornicu8:25:18 pmCASEWORKValentin Vornicu8:25:19 pmSince much of probability is based on counting (or at least the principles of counting), it should not be surprising that we must sometimes employ casework in probability.Valentin Vornicu8:25:25 pm

Valentin Vornicu8:25:33 pmA lot of this problem is in the wording! So let's make sure that we understand the problem.Valentin Vornicu8:25:38 pmWe are interested in rolls that are at least a five, or in other words rolls that are fives or sixes, so let's call a five or a six a "high roll," and everything else a "low roll." Which probability do we want to compute?Valentin Vornicu8:27:14 pmWe want to compute the probability that we roll at least five high rolls. In other words, we want to compute the probability that we roll five or six high rolls. How can we compute this probability?Valentin Vornicu8:29:19 pmWe can divide into the cases of rolling exactly five high rolls, and exactly six high rolls.Valentin Vornicu8:30:20 pmFirst, in how many different ways can we roll exactly five high rolls?Valentin Vornicu8:31:42 pmWe can roll exactly five high rolls in six ways. If there are exactly five high rolls, then there is exactly one low roll. We choose one of the six rolls to be the low roll, then all the other rolls must be high rolls.Valentin Vornicu8:31:45 pmWhat is the probability of rolling a high roll?Valentin Vornicu8:32:56 pmThe probability of rolling a high roll is 2/6 = 1/3, so the probability of rolling a low roll is 1 - 1/3 = 2/3. So what is the probability of rolling exactly five high rolls?Valentin Vornicu8:34:43 pm

Valentin Vornicu8:35:36 pmWhat is the probability of rolling exactly six high rolls?Valentin Vornicu8:36:31 pm

Valentin Vornicu8:36:52 pmSo what is the probability of rolling at least five high rolls?Valentin Vornicu8:37:26 pmThe probability of rolling at least five high rolls is 4/243 + 1/729 = 13/729. The answer is (A).Valentin Vornicu8:37:39 pm

Valentin Vornicu8:39:00 pmHow can we start on this problem?Valentin Vornicu8:40:53 pmWe can divide into cases. Which cases can we divide into?Valentin Vornicu8:41:23 pmWe can divide into cases based on the number of red faces.Valentin Vornicu8:41:27 pmLet n be the number of faces that are red, so n must be between 0 and 6. What is the probability that there are exactly n red faces?Valentin Vornicu8:43:33 pmWe can choose n of the six faces to be red in 6 choose n ways. The probability of each face being either color is 1/2.Valentin Vornicu8:43:38 pm

Valentin Vornicu8:45:01 pmNow let's go through the cases. Given that there are 0 red faces, what is the probability that the cube can be placed on the surface, so that all four vertical faces are all the same color?Valentin Vornicu8:45:57 pmIf there are 0 red faces, then all faces are blue, so the probability is 1.Valentin Vornicu8:46:04 pmGiven that there is one red face, what is the probability that the cube can be placed on the surface?Valentin Vornicu8:47:06 pmIf there is only one red face, then again the probability is 1.Valentin Vornicu8:47:08 pmGiven that there are two red faces, what is the probability that the cube can be placed on the surface?Valentin Vornicu8:49:20 pmIf we paint one face of the cube red, then there are five other faces to choose from for the other red face.Valentin Vornicu8:49:24 pmOnly face is opposite, so the probability that the two red faces are opposite each other is 1/5. This is also the probability that the cube can be placed on the surface.Valentin Vornicu8:49:27 pmGiven that there are three red faces, what is the probability that the cube can be placed on the surface?Valentin Vornicu8:50:13 pmTo place the cube on the surface, we must have at least four faces with the same color. There are only three red faces and three blue faces, so the probability is 0.Valentin Vornicu8:50:41 pmGiven that there are four red faces, what is the probability that the cube can be placed on the surface?Valentin Vornicu8:51:19 pmThis is the same as for the case of two red faces, because we can flip the colors.Valentin Vornicu8:51:22 pmHence, the probability in this case is also 1/5.Valentin Vornicu8:51:25 pmSimilarly, the probabilities for five red faces and six red faces are also 1.Valentin Vornicu8:51:27 pmHere's a summary of what we have found.Valentin Vornicu8:51:32 pm

Valentin Vornicu8:51:37 pmWhat do we want to do with these probabilities?Valentin Vornicu8:52:53 pmWe want to multiply the probabilities in each row, then add the products.Valentin Vornicu8:52:59 pm

Valentin Vornicu8:53:44 pmWhat does this simplify to?Valentin Vornicu8:53:53 pmThis simplifies to (1 + 6 + 3 + 3 + 6 + 1)/64 = 20/64 = 5/16. The answer is (B).Valentin Vornicu8:53:59 pmGEOMETRIC PROBABILITYValentin Vornicu8:54:05 pmAll the problems we have seen so far involve quantities that we can count, like the number of rolls of a die. But what if we have a probability problem involving a continuous quantity, like choosing a number from an interval? In these problems, we must take a geometric point of view.Valentin Vornicu8:54:12 pm

Valentin Vornicu8:54:38 pmLet the two real numbers be x and y. When is their product xy greater than zero?Valentin Vornicu8:56:24 pmThe product xy is greater than zero if and only if both x and y are greater than zero, or both x and y are less than zero.Valentin Vornicu8:56:30 pmWhat is the probability that a number chosen from the interval [-20,10] is greater than zero?Valentin Vornicu8:57:33 pmThe portion of the interval [-20,10] that is greater than 0 is (0,10]. The length of the interval (0,10] is 10, and the length of the interval [-20,10] is 30, so the probability of choosing a number greater than zero is 10/30 = 1/3.Valentin Vornicu8:58:20 pmIt follows that the probability of choosing a number less than zero is 1 - 1/3 = 2/3.Valentin Vornicu8:58:24 pmWhat is the probability that both x and y are positive?Valentin Vornicu8:59:52 pm

Valentin Vornicu9:00:19 pmWhat is the probability that both x and y are negative?Valentin Vornicu9:00:29 pm

Valentin Vornicu9:00:30 pmSo what is the probability that the product xy is greater than zero?Valentin Vornicu9:01:19 pm

Valentin Vornicu9:01:57 pmSUMMARYValentin Vornicu9:02:00 pmWhen solving a probability problem, the first step should be to determine what kind of probability you are dealing with. In some cases, all you have to do is compute the number of "successful" outcomes, and divide by the total number of outcomes. However, in other cases, you may have to use other techniques, such as tree analysis or geometric probability.Valentin Vornicu9:02:03 pmAlso, make sure you read the problem carefully. In probability, simply relying on intuition can easily lead to incorrect answers. Therefore, you should make sure you understand why your steps are correct.