1. A deck of cards is dealt out. What is the probability that the 14th card dealt is an ace?

What is the probability that the first ace occurs on the 14th card?

Solution:

The fourteenth card is equally likely to be any of the 52 cards. So the probability is 4/52.

Another way to see this is to look at the 52! arrangements of cards, out of these there

are 4.51! arrangements containing an ace as the 14 card. So the probability is

4 ×51!

52!=

4

52

The first ace occurs on the fourteenth card means that we only have 48 cards to choose

from in the first card followed by 47 for the second, …, then we have 4 ways to get an

ace in the out of 39 cards for the 14th card. So the probability is 48 × 47 × ⋯ × 36 × 4

52 × 51 × ⋯ × 40 × 39

2. A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the

probability that there will be

(a) no complete pair?

(b) exactly 1 complete pair?

Solution:

(a) There are a total of (20 C 8) ways of choosing 8 shoes. To have no pairs the selection

must have 8 different colors. There are (10 C 8) choices for color and two shoes for each

color.

𝐶108

𝐶208

× ( 𝐶21

)8

(b) in this case we need to choose one color for the pair and the remaining 6 shoes

must come from six different pairs.

𝐶101

× 𝐶96

𝐶208

× ( 𝐶21

)6

3. An elementary school is offering 3 language classes: one in Spanish, one in French, and

one in German. These classes are open to any of the 100 students in the school. There

are 28 students in the Spanish class, 26 in the French class, and 16 in the German class.

There are 12 students that are in both Spanish and French, 4 that are in both French and

German, and 6 that are in both Spanish and German, In addition, there are 2 students

taking all 3 classes. Let S be the event that a student is in Spanish, let F be the event that

a student is in French and let G be the event that a student is in German.

(a) If a student is chosen randomly, what is the probability that he or she is not in any

of these classes?

(b) If 2 students are chosen randomly, what is the probability that at least 1 is taking a

language class?

Solution:

(a) “Not in any of these classes” means “not in Spanish AND not in French AND

not in German,” or, 𝑃(𝑆𝑐 ∩ 𝐹𝑐 ∩ 𝐺𝑐). By De Morgan’s laws, this is the same as finding

𝑃 ((𝑆 ∪ 𝐹 ∪ 𝐺)𝑐) = 1 − 𝑃(𝑆 ∪ 𝐹 ∪ 𝐺) 𝑃(𝑆 ∪ 𝐹 ∪ 𝐺) = 𝑃(𝑆) + 𝑃(𝐹) + 𝑃(𝐺)– 𝑃(𝑆 ∩ 𝐹)– 𝑃(𝑆 ∩ 𝐺)

− 𝑃(𝐹 ∩ 𝐺) + 𝑃(𝑆 ∩ 𝐹 ∩ 𝐺)

= 0.28 + 0.26 + 0.16 − 0.12 − 0.06 − 0.04 + 0.02

= 0.50

1 − 𝑃(𝑆 ∪ 𝐹 ∪ 𝐺) = 1 − 0.5 = 0.5

(b) P(at least 1 is taking a language class) = 1 − P(neither is taking a language class)

1 −𝐶50

2

𝐶1002

4. If there are 12 strangers in a room, what is the probability that no two of them celebrate

their birthday in the same month?

Solution:

First, let’s find the size of the sample space. There are 12 months of the year and each of

the people can initially have a birthday in any one of these 12 months. The size of the

sample space is 1212

Now we impose the restriction that none of the 12 strangers has a birthday in the same

month. Let’s start with the first person. The first person can have a birthday in any of

the 12 months. Once this month is “chosen”, the next person cannot have a birthday in

the same month. The second person can have a birthday in any one of 11 months and

the third in any of 10 months and so on. Therefore, there are 12! ways for 12 strangers

to have a birthday in a different month. So,

P(each has a different birth month) = 12!

1212

5. An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected,

what is the probability that each of the balls will be

(a) of the same color

(i) Under sampling with replacement

(ii) Under sampling without replacement

(b) of different colors

(i) Under sampling without replacement

(ii) Under sampling with replacement

Solution:

(a) (i) Under sampling with replacement, we have the following. Each time the ball is

drawn, it is replaced and can be drawn again

(5

19)

3

+ (6

19)

3

+ (8

19)

3

= 0.124

(ii) . Of the 19 balls, we want to draw 3 of them so sample space is (19 C 3). we want to

draw 3 of the red balls, or 3 of the green balls, or 3 of the blue balls:

𝐶53

+ 𝐶63

+ 𝐶83

𝐶193

= 0.089

(b) (i) We are interested in drawing 1 red ball, 1 green ball and 1 blue ball:

𝐶51

+ 𝐶61

+ 𝐶81

𝐶193

= 0.248

(ii) 5

19×

6

19×

8

19= 0.035

6. You are dealt two cards from a well-shuffled deck. What is the probability that both are

of the same suit given that the first card is a spade? Do the problem directly and then do

it again using conditional probability.

Solution:

Doing the problem directly.

After the first card is dealt (which is a spade), there are 51 left in the deck. Of these, 12

are spades. So the probability is 12/51.

Using the definition of conditional probability:

Let T be the event of getting two cards of the same suit.

Let F be the event that the first card is a spade.

How many ways are there to get a sequence of two cards of the same suit, with the first

card being a spade?

That’s the same as both cards being spades. So 13 × 12.

The number of ways that the first card is a spade and the second could be any suit is

13× 51.

𝑃(𝑇|𝐹) =𝑃(𝑇 ∩ 𝐹)

𝑃(𝐹)=

13 × 12

13 × 51=

12

51

7. A simplified model for the movement of the price of a stock supposes that on each day

the stock’s price either moves up 1 unit with probability p or moves down 1 unit with

probability 1 − p. The changes on different days are assumed to be independent.

(a) What is the probability that after 2 days the stock will be at its original price?

(b) What is the probability that after 3 days the stock’s price will have increased by 1

unit?

(c) Given that after 3 days the stock’s price has increase by 1 unit, what is the probability

that it went up on the first day?

Solution:

(a) Let a string of k letters denote what the stock did the first k days, with U denoting

up and D denoting down. For the stock to return to its original price, we either have U D

or DU , each of which occurs with probability p(1 − p). Thus the probability that the

stock is at its original price after two days is 2p(1 − p).

(b) For the stock to increase 1 unit in 3 days, we either have U U D, U DU ,

or DU U , each of which occurs with probability p2(1 − p). Thus the

probability that the stock is at its original price after two days is

3p2(1 − p).

(c) We want

𝑃(𝑈 on first day|Increase 1 unit) =𝑃(𝑈 on first day AND increase 1 unit)

𝑃(Increase 1 unit)

=2𝑝2(1 − 𝑝)

3𝑝3(1 − 𝑝)=

2

3

8. Two cards are randomly chosen without replacement from an ordinary deck of 52 cards.

Let B be the event that both cards are aces, let As be the event that the ace of spades is

chosen, and let A be the event that at least one ace is chosen.

Find

(a) P(B|As)

(b) P(B|A)

Solution:

(a)

𝑃(𝐵|𝐴′𝑠) =𝑃(𝐵 ∩ 𝐴′𝑠)

𝑃(𝐴′𝑠)

B ∩ As is the event that both of the cards are aces and one of them is the ace of spades. There are 3 ways for this to happen, an ace of spades and an ace of diamonds, an ace of spades and an ace of hearts or an ace of spades and an ace of clubs. There are (52 C 2) ways of drawing 2 cards from a deck. The event As is the event that an ace of spades occurs. This can happen in one of 51 ways: an ace of spades and any other card in the deck.

=

3

𝐶522

51

𝐶522

=3

51

(b)

𝑃(𝐵|𝐴) =𝑃(𝐵 ∩ 𝐴)

𝑃(𝐴)

A ∩ B is the event that at least one ace is chosen and both cards are aces which is redundant: A ∩ B = B. P (B) Is the probability that both cards are aces. Of the 4 aces in the deck, we must choose 2 of them. P (A) is the probability that at least one of the cards is an ace. We use the complement; P (A) = 1 − P (neither card is an ace). To find P (neither card is an ace), of the 48 non- aces, we want to choose 2 of them and of the 4 aces, we want to choose none of them

=

𝐶42

𝐶522

1 −𝐶48

2

𝐶522

= 0.0303

9. Consider 3 urns. Urn A contains 2 white and 4 red balls; urn B contains 8 white and 4 red

balls; and urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what

is the probability that the ball chosen from urn A was white, given that exactly 2 white

balls were selected?

Solution:

𝑃(Ball from A white|2 white balls selected)

=𝑃(Ball from A white ∩ 2 white balls selected)

𝑃(2 white balls selected)

=(

26 ×

812 ×

34) + (

14 ×

412)

(26 ×

812 ×

34) + (

46 ×

812 ×

26) + (

26 ×

812 ×

14)

10. A recent college graduate is planning to take the first three actuarial examinations in the

coming summer. She will take the first actuarial exam in June. If she passes that exam,

then she will take the second exam in July, and if she also passes that one, then she will

take the third exam in September. If she fails an exam, then she is not allowed to take

any others. The probability that she passes the first exam is .9. If she passes the first

exam, then the conditional probability that she passes the second one is .8, and if she

passes both the first and the second exams, then the conditional probability that she

passes the third exam is .7.

(a) What is the probability that she passes all three exams?

(b) Given that she did not pass all three exams, what is the conditional probability that

she failed the second exam?

Solution:

(a)

𝑃(𝑝𝑎𝑠𝑠𝑒𝑠 𝑎𝑙𝑙 𝑒𝑥𝑎𝑚𝑠) = 𝑃(𝐸3𝐸2𝐸1) = 𝑃(𝐸3|𝐸2𝐸1) × 𝑃(𝐸2|𝐸1) × 𝑃(𝐸1) 𝑃(𝑝𝑎𝑠𝑠𝑒𝑠 𝑎𝑙𝑙 𝑒𝑥𝑎𝑚𝑠) = 𝑃(𝐸3𝐸2𝐸1) = 0.7 × 0.8 × 0.9

(b) Note that the event that she failed the second exam is 𝐸1𝐸2𝑐 and the event that she did

not pass all three exams is (𝐸1𝐸2𝐸3)𝑐

𝑃(𝐸1𝐸2𝑐|(𝐸1𝐸2𝐸3)𝑐) =

𝑃(𝐸1𝐸2𝑐 ∩ (𝐸1𝐸2𝐸3)𝑐)

𝑃((𝐸1𝐸2𝐸3)𝑐)=

(1 − 0.8) × 0.9

1 − 0.504≈ 0.3629