Probability exercises.docx

7/30/2019 Probability exercises.docx

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Probability exercises

I.Three horses A, B, and C are in a race. A is twice as likely to win as B, and

B is three times as likely to win as C. What's their respective probability ofwinning?

II.A gambler loads a die so that the probability of turning up a number ofpoints is directly proportional to the number of points (i.e., 2Pr(1)=Pr(2),etc.). Find Pr(6).

III.Flip a coin loaded so that Pr(H)=2Pr(T). If you get H, then randomly

choose a number between 1 and 8 (included). If you get T, then randomlychoose a number between 1 and 5 (included). What's the probability thatyou choose an even number?

IV.John is given two vaccines, A and B. They are for the same disease andact independently. The probability that A is successfully is 20%; thatprobability that B is successful is 40%. One successful vaccine is enoughfor vaccination. What's the probability that John is successfully vaccinated?

V.You are given two urns, A and B. A contains 4 red marbles, 2 whitemarbles, and 4 blue marbles. B contains 2 red marbles and 3 whitemarbles. Toss a fair die; if 2 or 5 appear, then choose a marble from B,otherwise choose a marble from A. What's the probability that youchoose:1. a red marble2. a white marble3. a blue marble

VI.You have a box with three coins, one fair, one two headed, and one suchthat Pr(T)=2/3. Randomly choose a coin from the box and flip it. What'sPr(H)? What's Pr(T)?


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VII.You have a bag with 1 red marble and 3 black marbles; you also have a faircoin and a coin loaded so that Pr(H)=3Pr(T). Choose a marblerandomly. If it's red, then flip the fair coin; if it's black, flip the loadedcoin. If you get H, then flip the other coin; if you get T, flip the samecoin. Determine:1. Pr(H on second flip)2. Pr(T on both flips)3. Pr(H on both flips)

Answers:

I.We know that Pr(A)=2Pr(B), and Pr(B)=3Pr(C); hence, Pr(A)=6Pr(C). But

Pr(A)+Pr(B)+Pr(C)=1. Consequently, 6Pr(C)+3Pr(C)+Pr(C)=1. So,Pr(C)=1/10; Pr(B)=3/10; Pr(A)=6/10.

II.We know that Pr(1)+Pr(2)+Pr(3)+Pr(4)+Pr(5)=Pr(6)=1; hence,Pr(1)+2Pr(1)+3Pr(1)+4Pr(1)+5Pr(1)+6Pr(1)=1. So,Pr(1)=1/21. Consequently, Pr(6)=6/21.

III.

Let's construct the tree:

Pr(E)=(2/3)x(1/2) + (1/3)x(2/5)= 14/30.


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IV.

Let's construct the tree, where "vaccine A is successful" is As and "vaccineB is successful" is Bs:

Pr(Joe is successfully vaccinated)=Pr(As v Bs)=1/5 + (4/5)x(2/5) = 13/25.Another approach is:Pr(Joe is successfully vaccinated)=Pr(As v Bs)= 1-Pr(-As & -Bs)= 1-[(4/5)x(3/5)]= 13/25

V.


Pr(R)= (1/3)x(2/5) + (2/3)x(4/10) = 6/15.Pr(W)= (1/3)x(3/5) + (2/3)x(2/10) = 1/3.Pr(B)= (2/3)x(4/10) = 4/15.


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VI.


Pr(H)= (1/3)x(1/2)+(1/3)+(1/3)x(1/3) = 11/18.Pr(T)= 1-Pr(H) = 1-(11/18) = 7/18.

VII.



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1. Pr(H on second flip) = (1/4)x(1/2)x(3/4) + (1/4)x(1/2)x(1/2) +(3/4)x(3/4)x(1/2) + (3/4)x(1/4)x(3/4) = 37/64.2. Pr(T on first flip & T on second flip)= (1/4)x(1/2)x(1/2) + (3/4)x(1/4)x(1/4)= 7/64.3. Pr(H on first flip & H on second flip) = (1/4)x(1/2)x(3/4) + (3/4)x(3/4)x(1/2)= 3/8.


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Introduction

Probability is a straight forward topic at school. There are only a few

areas of interest at this level. First though, you need to remember someof the fundamentals:

Probability ranges from 0 (impossible) to 1 (will happen). You

will NEVER get a probability beyond this range. If you do, your working is

wrong somewhere.

In probability, there are 3 classic examples - a coin, a die and a pack ofplaying cards.

1. The Coin

A coin has 2 possible outcomes --> HEADS or TAILS. In an unbiased coin,

each result is equally likely. So,

the probability of getting heads is "1 head out of 2 possible outcomes" --> P(heads) = 1/2

the probability of getting tails is "1 tail out of 2 possible outcomes" --

> P(tails) = 1/2

note: all the possible answers add up to give 1

2. The Die

A die has 6 possible outcomes --> 1,2,3,4,5 or 6. In an unbiased die, each

result is equally likely. So,

the probability of getting a 1 is "the single 1 out of 6 possible outcomes" -

-> P(1) = 1/6


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the probability of getting a 2 is "the single 2 out of 6 possible outcomes"

--> P(2) = 1/6


--> P(3) = 1/6the probability of getting a 4 is "the single 4 out of 6 possible outcomes"

--> P(4) = 1/6


--> P(5) = 1/6


--> P(6) = 1/6

note: all the possible answers add up to give 1

3. The Pack of cards

A pack of cards has 52 cards. There are a variety of patterns we could

use, eg.

there are 26 red and 26 black --> half and half

there are 4 suits --> clubs, diamonds, hearts and spadesthere are 13 cards in each suit

there are 4 of each number of card eg "6 of clubs", "6 of diamonds", "6 of

hearts" and "6 of spades".

The probability of picking a heart at random is "there are 13 hearts out

of 52 cards" OR "there are 4 suits, one of which is hearts". Both give the

same answer --> P(hearts) = 1/4

EXAMPLES

Find the following probabilities, given that all items are standard and non-

biased. Any card picked is also returned and the pack is shuffled too


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after every choice.

1. probability of a head, tossing a coin.

2. probability of rolling a 4 on a die.

3. probability of picking a red card4. probability of picking an ace

5. probability of picking a diamond

6. probability of picking the "9 of clubs"

solutions:

1. P(head) = 1/2

2. P(4) = 1/63. P(red) = 26/52 --> 1/2

4. P(ace) = 4/52 --> 1/13

5. P(diamond) = 13/52 --> 1/4

6. P(9 of clubs) = 1/52

EXAMPLES

A bag contains 1 yellow, 3 red and 4 blue marbles. Picking one marble atrandom and then replacing it in the bag, find the following probabilities:

1. probability of picking the yellow marble.

2. probability of picking a red marble.

3. probability of picking a blue marble.

4. probability of not picking a red marble.

solutions:

1. P(yellow) = 1/8

2. P(red) = 3/8

3. P(blue) = 4/8 --> 1/2

4. P(not red)? well not red means it can be either yellow or blue. How

many is that? it is 1+4 marbles... ie 5 out of 8 marbles. So, P(not red) =


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5/8

The following is a very important fact to remember:P(event) + P(not event) = 1

This means that whatever the probability is for "an event to occur", the

probability of "the event not occuring" is enough to make both

probabilities add to 1. This is because the event happens or it doesn't

happen - together there is no other possible outcome, so together their

probabilities have to add up to 1.

so. going back to question 4, P(not red) is 1-P(red)

--> P(not red) = 1 - 3/8

--> P(not red) = 5/8

which is the same answer we got by thinking about the yellow and blue

marbles.

Conclusion

This is the end of the basics of Probability. If we stopped here, the topic

would be pretty useless in real world situations, so the next lesson will

advance the theory a little into such situations as:

"find the probabily of rolling a 2 or a 3 on a die"

or

"what is the probability of rolling a 2 and also picking an ace from a pack

of cards".


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Introduction to Probability

Probabilityis a way to describe thelikelihoodof something happening. You usually

encounterprobabilityoften without even realizing it. For example, when the weatherreport says that there is a 60% chance of rain today, that is anexpressionofprobability. And if someone says that you have a 50-50 chance of guessing a coin toss

- that too, is anexpressionof probability. While these are everyday occurrences that

deal with probability, when we talk aboutprobabilitymathematically, we usually writeprobabilities either infractionor decimal form.

In general, we say that theprobabilityof something happening is theratioof the

number of ways that thing can happen to the total number of ways for all things tohappen. The thing we want to happen is usually called the event. So we will need to

know the number of ways for the event to happen and the total number of ways for all

events to happen. In a simpler form,

For example, lets think about rolling a die (this is singular for dice). A die has six sidesand a number from 1 to 6 appears on each side. We usually assume that we have afair die meaning that each number has an equal chance of occurring.

Lets talk about theprobabilityof rolling a 4. In this case, rolling a 4 is the event weare interested in. There is only one way for this to happen, so 1 is the numerator of

the ratio. Because there are six possibilities, 6 will be the denominator of the ratio. We

usually express theprobabilityof rolling a 4 as:

The six possibilities that we use as the denominator of theratioare usually referred toas the sample space. The samplespaceis where you list all the possible outcomes

for an experiment. Our experiment was to roll the die one time. Sometimes listingthe samplespaceis very helpful in knowing how many outcomes there are in anexperiment. Other times, like with rolling a die, you can just think about how many

ways something can happen. And still other times, there may be so many possibilities

that you dont want to list them and you cant just think about them easily. We wontworry about those situations here, but be aware of all the ways you can deal with asamplespace(its simple and you dont have to list it, its helpful to list all the

possibilities, its too complicated to list every possibility).

It is interesting to note that theprobabilityof rolling any number 1 - 6 will always

be since all numbers have an equal chance of happening. It is also important to

note that if you add all the probabilities together, you will get 1.

When you add all the probabilities associated with all the events of an experiment, you
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will get one. This is an important rule to remember when working with probabilities.

Knowing that all the probabilities associated with anexperimentequal one can save

some time in working some problems.

What if a problem asked you to find theprobabilityof NOT rolling a 3? That means you

would be looking for theprobabilityof rolling a 1, 2, 4, 5, or 6. You would need to findall five of those individual probabilities and add them up. But an easier way is to workwith the complement. Thecomplement is a way of finding theprobabilityof an event

NOT happening. So theprobabilityof NOT getting a 3 is found by finding

theprobabilityof 3 and then subtracting from 1.

or .

Another important rule deals with the type of numbers that are acceptable as

aprobabilityanswer. Probabilities can only take on values from 0 to 1. Keep in mindthat 0 and 1 are acceptable values for aprobabilityanswer. Mathematically this is

represented as .

Aprobabilityof 0 means that an event is impossible and aprobabilityof 1 means

that an event is certain.

For example, if we go back to our die problem, theprobabilityof rolling a 7 is zero

because you can never roll a 7 with just one die.

Theprobabilitythat you are used a computer to access this lesson is 1 because theonly way to see these lessons is on-line or by printing the web page. Either way, it is

certain that you used a computer to access this lesson.

P(use a computer to access this lesson) = 1

If an event is neither certain nor impossible, then itsprobabilityshould be somewhere

between 0 and 1. If you perform a computation for aprobabilityand your answer isnegative or larger than 1, then your answer is incorrect. This will be useful in laterlessons as you perform more complicated computations.

In addition to the die examples already discussed there are several other commontypes of problems that you will come across when working with probability.

Common Examples:

i. Marbles in a Box

Suppose there are 3 red marbles, 6 blue marbles, and 7 green marbles in a box.Theprobabilityof pulling out a red marble is 3/16. There are 3 red marbles whichgives us our numerator of 3 and there are 16 total marbles which gives us ourdenominator.
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ii. Cards

In order to solve problems involving cards, you should learn some basic facts about a

deck of cards if you are not familiar with cards.

There are 52 cards in a deck. The 52 cards are broken up into 4 suites (spades, clubs, diamonds,

hearts).

Each suite has 13 cards (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10. Jack, Queen,King).

There are 4 of each type of card in the deck (4 aces, 3 10s, etc.)

Two suites are red (diamonds and hearts). Two suites are black (spades and clubs).

So a simpleprobabilityproblem that is very common would be to ask what is

theprobabilityof pulling out a 6. Thisprobabilitywould be 4/52 which could then besimplified to 1/13. Notice there are 4 sixs in a deck out of a total of 52 cards which ishow we get 4/52. You should always simplify your final answer when expressing aprobability.

iii. Dice

We have looked at some examples that deal with rolling one die. However it is very

common to see problems that involve rolling two dice.

When rolling two dice, there are 36 possibilities. It is usually helpful to consider a firstdie and a second die to keep the two distinct. The possibilities listed below are ordered

pairs indicating the number of the first die and then the number on the second die. To

find the sum, simply add the two numbers.(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

So if we want to find theprobabilityof rolling a sum of 8, we need to find all thepossible ways to roll an 8 (there are 5) out of the total possibilities which is 36.

Therefore, P(rolling sum of 8) = 5/36

Addition Rule of ProbabilityThere are many rules associated with solvingprobabilityproblems. This lesson deals

with the addition rule. The addition rule helps you solveprobabilityproblems thatinvolve two events. Even though we discuss two events (usually labeled A and B),were really talking about performing one task (rolling dice, drawing cards, spinning a

spinner, etc.) and finding about theprobabilityof two things happening in that onetask.
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When asked to find theprobabilityof A or B, wemeanthat A can happen, or B can

happen, or both can happen together. This is what is stated in the addition rule.

The Addition Rule:

Consider events A and B. P(A B)= P(A) + P(B) - P(A B)

What The Rule Means:

Suppose we roll two dice and want to find theprobabilityof rolling a sum of 6 or 8.

This can be written in words as P(6 or 8) or more mathematically is P(6 8).Remember that OR (the union symbol ) means that one or the other or both events

can happen. So what is theprobabilityof getting a 6 or an 8 or both? You may want torefer to the dicechartinintroductory lesson on basic probabilityif you need to

familiarize yourself with the outcomes of rolling two dice.

P(6) = 5/36P(8) = 5/36

P(6 and 8 together) is impossible so theprobabilityis 0.

So P(6 8) = 5/36 + 5/36 - 0 = 10/36 = 5/18

Since rolling a sum of 6 and 8 cannot happen together at the same time, we say thatthey are disjoint or mutually exclusive. When two events are disjoint, you do nothave to worry about subtracting theprobabilityof both events happening together

since thatprobabilitywill always be 0. Since it can sometimes save a step, a lot of

students like to determine if events are disjoint before finding individual probabilities

Let's Practice:

i. You are going to pull one card out of a deck. Find P(Ace King).

The addition rule says we need to find P(Ace) + P(King) - P(both). If you need tofamiliarize yourself with the features of a deck of cards, refer tointroductory lesson on

basic probabilityfor more information.

P(Ace) = 4/52P(King) = 4/52

P(both at the same time) = 0

P(Ace King) = 4/52 + 4/52 = 8/52 = 2/13

Notice that the final answer is always simplified. However, most of the time it is bestto NOT simplify along the way so that youll have easy common denominators.

ii. You are going to roll two dice. Find P(sum that is even or sum that is a multipleof 3).
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The addition rule says we need to find P(even) + P(multiple of 3) - P(both).

If you need to familiarize yourself with the possible combinations of rolling two dice,

refer tointroductory lesson on basic probabilityfor achartshowing all 36 possibilities.

P(even) means how many ways to roll 2, 4, 6, 8, 10, or 12.

P(even) = 18/36

P(multiple of 3) means how many ways to roll 3, 6, 9 or 12.

P(multiple of 3) = 12/36

P(both) means what is the overlap. Notice that 6 and 12 occur in both places and have

been counted twice. We need to subtract those out.P(both) = 6/36

So P(sum that is even or a multiple of 3) = 18/36 + 12/36 - 6/36 = 24/36 = 2/3.

Multiplication Rule of Probability

Theaddition rulehelped us solve problems when we performed one task and wantedto know theprobabilityof two things happening during that task. This lesson dealswith the multiplication rule. The multiplication rule also deals with two events, but in

these problems the events occur as a result of more than one task (rolling one die

then another, drawing two cards, spinning a spinner twice, pulling two marbles out ofa bag, etc).

When asked to find theprobabilityof A and B, we want to find out theprobabilityofevents A and B happening.

The Multiplication Rule:

Consider events A and B. P(A B)= P(A) P(B).

Note: Some books will say to take care that A and B are independent, but the rule canalso be used with dependent events, you just have to be more careful in find P(A) andP(B).

What The Rule Means:

Suppose we roll one die followed by another and want to find theprobabilityof rolling

a 4 on the first die and rolling an even number on the second die. Notice in thisproblem we are not dealing with the sum of both dice. We are only dealing withtheprobabilityof 4 on one die only and then, as a separate event, theprobabilityof

an even number on one die only.

P(4) = 1/6

P(even) = 3/6

So P(4 even) = (1/6)(3/6) = 3/36 = 1/12

While the rule can be applied regardless of dependence or independence of events, we

should note here that rolling a 4 on one die followed by rolling an even number on thesecond die are independent events. Each die is treated as a separate thing and what
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happens on the first die does not influence or effect what happens on the second die.

This is our basic definition of independent events: the outcome of one event does notinfluence or effect the outcome of another event.

Well look at examples later that deal withdependent events. Just keep in mind that

what happens on one event will effect the other event.

Let's Practice:

i. Suppose you have a box with 3 blue marbles, 2 red marbles, and 4 yellow

marbles. You are going to pull out one marble, record its color, put it back inthe box and draw another marble. What is theprobabilityof pulling out a redmarble followed by a blue marble?

The multiplication rule says we need to find P(red) P(blue).

P(red) = 2/9P(blue) = 3/9

P(red blue) = (2/9)(3/9) = 6/81 = 2/27

The events in this example were independent. Once the first marble was pulled out

and its color recorded, it was returned to the box. Therefore, theprobabilityfor thesecond marble was not effected by what happened on the first marble.

Notice that the final answer is always simplified. Some students find it helpful tosimplify before multiplying, but the final answer must always be simplified.

ii. Consider the same box of marbles as in the previous example. However in this

case, we are going to pull out the first marble, leave it out, and then pull out

another marble. What is theprobabilityof pulling out a red marble followed bya blue marble?

We can still use the multiplication rule which says we need to find P(red) P(blue).But be aware that in this case when we go to pull out the second marble, there will

only be 8 marbles left in the bag.

P(red) = 2/9

P(blue) = 3/8

P(red blue) = (2/9)(3/8) = 6/72 = 1/12

The events in this example were dependent. When the first marble was pulled out andkept out, it effected theprobabilityof the second event. This is what is meant by

dependent events.

iii. Suppose you are going to draw two cards from a standard deck. What is
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theprobabilitythat the first card is an ace and the second card is a jack (justone of several ways to get blackjack or 21).

Using the multiplication rule we get

P(ace) P(jack) = (4/52)(4/51) = 16/2652 = 4/663

Notice that this will be the sameprobabilityeven if the question had asked for

theprobabilityof a jack followed by an ace.
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