4204 words - 17 pages

21-110: Problem Solving in Recreational Mathematics

Homework assignment 7 solutions

Problem 1. An urn contains ﬁve red balls and three yellow balls. Two balls are drawn from the

urn at random, without replacement.

(a) In this scenario, what is the experiment? What is the sample space?

(b) What is the probability that the ﬁrst ball drawn is red?

(c) What is the probability that at least one of the two balls drawn is red?

(d) What is the (conditional) probability that the second ball drawn is red, given that the ﬁrst ball

drawn is red?

Solution.

(a) The experiment is the drawing of two balls from the urn without replacement. The sample

space is the set of possible outcomes, of which ...view middle of the document...

Each ball is equally likely to

be drawn. So the probability that the ﬁrst ball drawn is red is 5/8.

(c) The event that at least one of the two balls is red contains three outcomes: RR, RY, and YR.

Since we know the probabilities of all of these outcomes, we can ﬁnd the probability of this event

by adding the probabilities of the individual outcomes. So the probability that at least one of the

two balls is red is 5/14 + 15/56 + 15/56 = 25/28.

Alternatively, we can observe that the event that at least one of the two balls is red is the

complement of the event that both balls are yellow. The probability that both balls are yellow

is 3/28, so the probability that it is not the case that both balls are yellow (i.e., the probability that

at least one ball is red) is 1 − 3/28 = 25/28.

Page 1

(d) Suppose the ﬁrst ball drawn is red. Then there are seven remaining balls, and four of them

are red. So the conditional probability that the second ball is red, given that the ﬁrst ball is red,

is 4/7.

Problem 2. Suppose you and a friend play a game. Two standard, fair, six-sided dice are thrown,

and the numbers appearing on the dice are multiplied together. If this product is even, your friend

gives you a quarter, but if this product is odd, you must give your friend one dollar.

(a) What is the expected value of this game for you? Round to the nearest cent.

(b) What is the expected value of this game for your friend? Round to the nearest cent.

(c) Is this game fair? Why or why not?

Solution. We begin by drawing a table of the possible dice throws. The 36 outcomes represented

by the squares of the grid below are all equally likely. The shaded squares represent outcomes in

which the product of the numbers on the dice is even, and the unshaded squares represent outcomes

in which this product is odd.

5 6

1 2 3 4

1

1

2

3

4

5

2

2

4

6

8

10 12

3

3

6

9

12 15 18

4

4

8

12

16 20 24

5

5

10 15

20 25 30

6

6

12 18

24 30 36

6

(a) From the table above, we see that 27 of the 36 equally likely outcomes yield an even product, so

the probability that the product is even is 27/36, or 3/4. Similarly, the probability that the product

is odd is 9/36, or 1/4. You will win 25/ if the product of the numbers on the dice is even, but

c

you will lose $1 if the product is odd. So there is a 3/4 probability that you will gain 25/ and a

c

1/4 probability that you will lose 100/ (that is, gain −100/). So the expected value of the game for

c

c

you is

3

1

c

c

c

4 (25/) + 4 (−100/) = −6.25/.

Rounded to the nearest cent, this is −6/. This means that you should expect to lose about 6/, on

c

c

average, each time you play this game.

(b) Any money that you win in this game is money that your friend loses, and any money that

your friend wins is money that you lose. (Mathematicians call this a zero-sum game.) Therefore, if

the expected value of the game for you is...

Tap into the world’s largest open writing community