MATH Section 7.1

What is the probability that a five-card poker hand contains a flush (including straight and royal flushes), that is, five cards of the same suit? (Enter the value of probability in decimals. Round the answer to five decimal places.) Explanation Of the C(52, 5) = 2,598,960 hands, 4·C(13, 5) = 5148 are flushes, since a flush can be specified by choosing a suit and then choosing 5 cards from that suit. The probability is 5148/2598960=33/16660≈0.00198.

0.00198

What is the probability that a five-card poker hand contains the two of diamonds and the three of spades? (Enter the value of probability in decimals. Round the answer to four decimal places.) Explanation Compute the number of poker hands that contain the two of diamonds and the three of spades. There is no choice about choosing these two cards. To form the rest of the hand, choose 3 cards from the 50 remaining cards, so there are C(50, 3) hands containing these two specific cards. Hence, the probability is C(50,3)C(52,5)=5/663≈0.0075.

0.0075

What is the probability that a fair die comes up six when it is rolled? (Enter the value of probability in decimals. Round the answer to two decimal places.) Explanation There are six equally likely outcomes when a fair die is rolled. There is only one outcome for rolling a six when a fair die is rolled. The probability is 16≈0.17.

0.17

What is the probability that a card selected at random from a standard deck of 52 cards is an ace or a heart? (Enter the value of probability in decimals. Round the answer to two decimal places.) Explanation There are 13 hearts and 4 aces in a pack of cards. Among these one card will be an ace of hearts. By the inclusion-exclusion principle, there are 13 + 4 - 1 = 16 cards that are aces or hearts. The probability is 16/52=4/13≈0.31.

0.31

Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive integers not exceeding the given integers. (Enter the value of probability in decimals. Round the answer to two decimal places.) 40 Explanation If the numbers are chosen from the integers from 1 to n, then there are C(n, 6) possible entries. To avoid all the winning numbers, make the choice from the n − 6 nonwinning numbers. This can be done in C(n − 6, 6) ways. Since the winning numbers are picked at random, the probability is C(n − 6, 6)C(n, 6)C(n⁢ − 6, 6)C(n, 6) . The probability of selecting none of the correct six integers in a lottery from the positive integers not exceeding 40 is C(34, 6)C(40, 6) = 13449043838380 ≈ 0.35C(34, 6)C(40, 6)⁢ = 13449043838380⁢ ≈ 0.35 .

0.35

Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive integers not exceeding the given integers. (Enter the value of probability in decimals. Round the answer to two decimal places.) 48 Explanation If the numbers are chosen from the integers from 1 to n, then there are C(n, 6) possible entries. To avoid all the winning numbers, make the choice from the n − 6 nonwinning numbers. This can be done in C(n − 6, 6) ways. Therefore, since the winning numbers are picked at random, the probability is C(n − 6, 6)C(n, 6)C(n⁢ − 6, 6)C(n, 6) . The probability of selecting none of the correct six integers in a lottery from the positive integers not exceeding 48 is C(42, 6)C(48, 6) = 524578612271512 ≈ 0.43C(42, 6)C(48, 6)⁢ = 524578612271512⁢ ≈ 0.43 .

0.43

What is the probability that the sum of the numbers on two dice is even when they are rolled? (Enter the value of probability in decimal format.) Explanation There are 36 equally likely outcomes when two fair dice are rolled. (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) The sum of the numbers on two dice will be even if and only if the number on the second die has the same parity (even or odd) as the first. Since there are 3 even and 3 odd faces, whatever the parity of the number on the first die, exactly three out of six numbers of the second die will have the same parity and the sum will be even. So, there are 18 possible outcomes for the sum of the numbers to be even when two dice are rolled. The probability is 1836=12=0.51836=12=0.5 .

0.5

What is the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing if 50 people enter a contest and satisfying the following conditions? (Enter the value of probability in decimals. Round the answer to two decimal places.) Winning more than one prize is allowed. Explanation There are 50 · 50 · 50 · 50 equally likely outcomes of the drawings. The probability is 150 ⋅ 50 ⋅ 50 ⋅ 50=16250000 ≈ 1.60 × 10−7150 · 50 · 50 · 50=16250000⁢ ≈ 1.60⁢ × 10−7 .

1.60 × 10^-7

What is the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing if 50 people enter a contest and satisfying the following conditions? (Enter the value of probability in decimals. Round the answer to two decimal places.) No one can win more than one prize. Explanation There are 50 · 49 · 48 · 47 equally likely outcomes of the drawings. The probability is 150 ⋅ 49 ⋅ 48 ⋅ 47=15527200 ≈ 1.81 × 10− 7150 · 49 · 48 · 47=15527200⁢ ≈ 1.81⁢ × 10− 7 .

1.81 × 10^-7

What is the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing if 52 people enter a contest and no one can win more than one prize? Explanation It is given that no one can win more than one prize. Thus, in the first draw, all 52 people have equal chances of winning a prize. In the next draw, only 51 people have equal chances of winning a prize as the person who wins the first draw cannot win another prize. Similarly, in the third and fourth draws, 50 and 49 people, respectively, have equal chances of winning. Thus, there are 52 · 51 · 50 · 49 equally likely outcomes of the drawings. In only one of these draws do Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively. Therefore, the probability is 152 ⋅151 ⋅150 ⋅149 =16,497,400.152 ·151 ·150 ·149 =16,497,400.

1/6,497,400

What is the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing if 52 people enter a contest and winning more than one prize is allowed? Explanation Since winning more than one prize is allowed, in each draw, 52 people have equal chances of winning a prize. Thus, there are 52 · 52 · 52 · 52 equally likely outcomes of the drawings. In only one of these draws do Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively. Therefore, the probability is 152⋅152⋅152⋅152=17,311,616.152·152·152·152=17,311,616.

1/7,311,616

Find the probability of winning a lottery by selecting the correct six integers, where the order in which these integers are selected does not matter, from the positive integers not exceeding the given integers. (Enter the value of probability in decimals. Round the answer to one decimal place.) The probability of winning a lottery by selecting the correct six integers from the positive integers not exceeding 50 is _____ × 10-8. Explanation If the numbers are chosen from the integers from 1 to n, then there are C(n, 6) possible entries, only one of which is the winning one. So the answer is 1/C(50, 6) = 1/15890700 ≈ 6.3 × 10^−8

6.3

What is the probability that a five-card poker hand contains the ace of hearts? (Enter the value of probability in decimals. Round the answer to one decimal place.) Explanation There are C(52, 5) possible poker hands, and by symmetry, they are all equally likely. In order to solve this problem, compute the number of poker hands that contain the ace of hearts. To form the rest of the hand, choose 4 cards from the 51 remaining cards, so there are C(51, 4) hands containing the ace of hearts. Hence, the probability is C(51,4)C(52,5)=5/52≈9.6%.

9.6%

Two events E1 and E2 are called independent if p( E1 ∩∩ E2) = p(E1)p(E2). For each of the following pairs of events, which are subsets of the set of all possible outcomes when a coin is tossed three times, determine whether or not they are independent. Let E1 be the event that tails come up when the coin is tossed the first time and E2 be the event that heads come up when the coin is tossed the second time. Drag the probability values from the right column and drop them in the corresponding events in the left column. Explanation Intuitively, these should be independent, since the first event seems to have no influence on the second. In fact we can compute as follows: First p(E1) = 1/2 and p(E2) = 1/2 by the symmetry of coin tossing. Furthermore, E1 ∩ E2 is the event that the first two coins come up tails and heads, respectively. Since there are four equally likely outcomes for the first two coins (HH, HT, TH, and TT ), p(E1 ∩ E2) = 1/4.

p(E₁) and p(E₂): 1/2 p(E₁ ∩ E₂): 1/4 Independence: Independent