Probability and Statistics Problems and Solutions, Exercises of Mathematics

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MA 110-

§3.1 – 4.

Test

score

Name:

13 April 2002

1. Two dice are rolled. Write out the complete sample space and use it to find the probability

that two dice differ by more than 2. Explain. (10 points)

Solution: (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)

p( dice differ by more than 2 ) = 1236 ≈ 0_._ 33

2. Four fair coins are tossed. Find the probability that there are more heads than tails in the

result. Explain. (10 points)

Solution: For there to be more heads than tails, there would have to be 3 heads and 1 tail, or all heads. The are 4ways the former can occur and only 1 way the latter can occur. So out of the 24 = 16 outcomes of tossing 4coins, p( more heads than tails ) = 165 ≈ 0_._ 31.

3. Find the probability of being dealt four cards from the same kind in a five-card hand.

Explain your counting methods. Express your answer as a decimal number. (10 points)

Solution: First, choose a kind ( 13 C 4 ), then choose all 4cards from the selected kind ( 4 C 4 ), then select the fifth card from the 48 remaining ( 48 C 1 ). So,

p( 4-of-a-kind ) = 13

C 1 · 4 C 4 · 48 C 1

52 C 5

=

13 · 1 · 48

2 , 598 , 960

≈ 0. 00024

4. If three dice are rolled, find the probability that there is at least one six among the three

results. Explain. (10 points)

Solution: We count the complementary event where we have no sixes.

p( at least one six ) = 1 − p( no six ) = 1 −

= 1 −

=

≈ 0. 42

5. We play a lottery in which four numbers in the range 1 through 15 are selected. Find the

probability of winning this lottery, i.e., the probability of picking the four correct numbers.

Then find the probability of picking exactly three of the four correct numbers. Explain.

(10 points)

Solution: p( all four winning numbers ) =

15 C 5

=

≈ 0. 00073

p( three of the four winning numbers ) = 4

C 3 · 11 C 1

15 C 5

=

≈ 0. 0032

MA 110 Test 2 page 2

6. You and two of your best math friends decide to play a game. Each of you flips a coin

simultaneously. If all three coins match, each of your friends pays you $5. If they don’t all

match, you pay each of your friends $2. What is the expected value of this game from your

point of view? Would this be a profitable game for you to play repeatedly? (10 points)

Solution: The expected value is

p( all heads or all tails ) · ( $10 ) + p( both heads and tails occur ) · ( −$4 ) = −$0_._ 50

So the game is not in your favor; you should expect to lose an average of $0_._ 50 per game.

7. Two cards are selected. Let E 1 denote the event that the first card is a heart, and let E 2

denote the event that the second card is a heart. Are E 1 and E 2 mutually exclusive? Are

they independent? Explain. (10 points)

Solution: The events are not mutually exclusive since E 1 ∩ E 2 contains the outcomes with a heart on both selections. The events are not mutually exclusive either since p(E 2 ) = 1352 whereas p(E 2 | E 1 ) = 1251.

8. An electronics manufacturer buys its chips from three factories: A, B, and C. Factory A

supplies half of the total chips while B and C each supply one-quarter. If the defective rates

of the chips from the three factories are 1%, 2%, and 3%, respectively, find the probability

that a randomly selected chip is defective. (10 points)

Solution: Let Ex denote the event that a chip was manufactured at factory x and let E denote the event that a chip is defective. Then p(E) = p(E | EA ) · p(EA ) + p(E | EB ) · p(EB ) + p(E | EC ) · p(EC ) = 0_._ 50 · 0_._ 01 + 0_._ 25 · 0_._ 02 + 0_._ 25 · 0_._ 03 = 0_._ 0175

9. Draw a relative frequency density histogram

for the dataset {3.8, 4.5, 5.3, 5.6, 6.2,6.6, 7.2,

7.4, 8.9, 7.1, 9.2, 8.2, 7.7, 7.9, 8.1, 7.9, 8.1, 9.4,

9.9}. Use 5 data groups (bins): 0 ≤ x < 5,

5 ≤ x < 7, 7 ≤ x < 8, 8 ≤ x < 9, and

9 ≤ x ≤ 10. (10 points)

Solution: category freq. relative freq. density 0 ≤ x < (^5 2 192192) · 5 ≈ 0_._ 02 5 ≤ x < (^7 4 194194) · 2 ≈ 0_._ 11 7 ≤ x < (^8 6 196196) · 1 ≈ 0_._ 32 8 ≤ x < (^9 4 194194) · 1 ≈ 0_._ 21 9 ≤ x ≤ (^10 3 193193) · 1 ≈ 0_._ 16 Total 19

10. Calculate the mean and median of the data set S = { 8 , 3 , 12 , 17 , 11 , 14 , 9 , 7 , 5 , 13 } (10 points)

Solution: Order the data we get 3, 5, 7, 8, 9, 11, 12, 13, 14, 17. So the median of the 10 data points is halfway between items 5 and 6: 9 + 211 = 10. The mean is given by

x ¯ = Σ xi 10

= 99

= 9. 9

.