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Probability Theory Homework Solutions: Independent Events, Combinatorics, and Dice Rolls, Assignments of Mathematics

University of South Carolina (USC) - Columbia Mathematics

Solutions to various probability theory problems involving independent events, combinatorics, and dice rolls. Topics include calculating probabilities of intersections and unions of events, finding the probability of exactly one or two successful events, and determining the probability of at least one match between rolls and assigned numbers. The document also covers finding the probability of winning a game with replacement or no replacement, and finding the probability of selecting a specific type of bean or coin.

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Uploaded on 10/01/2009

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MATH 511, Meade HW Solutions

2.4 – 3, 7,17, 10, 16 2/13/04

3. P(A) = 1/4 , P(B) = 2/3 , A and B are independent

a. P(A ∩ B) = (1/4) (2/3) = 1/6

b. P(A ∩ B’) = (1/4) (1 - 2/3) = (1/4) (1/3) = 1/12

c. P(A’ ∩ B’) = (1- 1/4 ) (1 – 2/3) = (3/4) (1/3) =1/4

d. P[ (A U B)’] = 1 – P(A U B) = 1 – (P(A) + P(B) – P(A ∩ B)

= 1 – (1/4) – (2/3) + (1/6) = 1/4

e. P(A’ ∩ B) = (1 – 1/4 ) (2/3) = (3/4) (2/3) = 1/2

7. P(A1) = 0.5, P(A2) = 0.7, P(A3) = 0.6, events are mutually independent

a. Compute the probability that exactly one player scores a field goal.

This means that one player succeeds while the other two fail. There are

three ways that this can happen.

P(A1) P(A2’) P(A3’) + P(A1’) P(A2) P(A3’) + P(A1’) P(A2’) P(A3) =

(0.5)(0.3)(0.4) + (0.5)(0.7)(0.4) + (0.5)(0.3)(0.6) = 0.06 + 0.14 + 0.09 =

0.29

b. Compute the probability that two are successful.

P(A1) P(A2) P(A3’) + P(A1) P(A2’) P(A3) + P(A1’) P(A2) P(A3) =

(0.5)(0.7)(0.4) + (0.5)(0.3)(0.6) + (0.5)(0.7)(0.6) = 0.14 + 0.09 + 0.21 =

0.44

17. Each of 12 students is assigned a number 1-12 and is given a 12-sided die

a. What is the probability of one at least match between what the students

roll and their assigned number?

This is the complement of none of the students getting a match.

The probability that none of them gets a match is (11/12)12 since each

student has an 11/12 chance of not matching. So the probability of no one

matching is: 1 – (11/12)12 = 64.8%

b. If you are one of the students, what is the probability that at least one

student matches the number that you roll?

The is the complement that no one matches you. The chance of someone

not rolling the same number as you is 11/12. The chance that no one

matches you is (11/12)11 and the chance that at least one matches you is

1 - (11/12)11 = 61.6%

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MATH 511, Meade HW Solutions 2.4 – 3, 7,17, 10, 16 2/13/

P(A) = 1/4 , P(B) = 2/3 , A and B are independent a. P(A ∩ B) = (1/4) (2/3) = 1/ b. P(A ∩ B’) = (1/4) (1 - 2/3) = (1/4) (1/3) = 1/ c. P(A’ ∩ B’) = (1- 1/4 ) (1 – 2/3) = (3/4) (1/3) = 1/ d. P[ (A U B)’] = 1 – P(A U B) = 1 – (P(A) + P(B) – P(A ∩ B) = 1 – (1/4) – (2/3) + (1/6) = 1/ e. P(A’ ∩ B) = (1 – 1/4 ) (2/3) = (3/4) (2/3) = 1/
P(A 1 ) = 0.5, P(A 2 ) = 0.7, P(A 3 ) = 0.6, events are mutually independent a. Compute the probability that exactly one player scores a field goal. This means that one player succeeds while the other two fail. There are three ways that this can happen. P(A 1 ) P(A 2 ’) P(A 3 ’) + P(A 1 ’) P(A 2 ) P(A 3 ’) + P(A 1 ’) P(A 2 ’) P(A 3 ) = (0.5)(0.3)(0.4) + (0.5)(0.7)(0.4) + (0.5)(0.3)(0.6) = 0.06 + 0.14 + 0.09 = 0. b. Compute the probability that two are successful. P(A 1 ) P(A 2 ) P(A 3 ’) + P(A 1 ) P(A 2 ’) P(A 3 ) + P(A 1 ’) P(A 2 ) P(A 3 ) = (0.5)(0.7)(0.4) + (0.5)(0.3)(0.6) + (0.5)(0.7)(0.6) = 0.14 + 0.09 + 0.21 = 0.
Each of 12 students is assigned a number 1-12 and is given a 12-sided die a. What is the probability of one at least match between what the students roll and their assigned number? This is the complement of none of the students getting a match. The probability that none of them gets a match is (11/12)^12 since each student has an 11/12 chance of not matching. So the probability of no one matching is: 1 – (11/12)^12 = 64.8% b. If you are one of the students, what is the probability that at least one student matches the number that you roll? The is the complement that no one matches you. The chance of someone not rolling the same number as you is 11/12. The chance that no one matches you is (11/12)^11 and the chance that at least one matches you is 1 - (11/12)^11 = 61.6%

Let D 1 , D 2 , D 3 all be four sided die. They are labeled as follows: D 1 : 0, 3, 3, 3 D 2 : 2, 2, 2, 5 D 3 : 1, 1, 4, 6 a. Find the probability that D 1 beats D 2. (3)(3)/ (4)(4) = 9/ b. Find the probability that D 2 beats D 3. [(3)(2) + (1)(3)]/ (4)(4) = 9/ c. Find the probability that D 3 beats D 1. [(2)(4) + (2)(1)]/ (4)(4) = 10/
You and an opponent take turns drawing from five balls. Four are marked LOSE and one is marked WIN. The first to draw the WIN ball wins. If you Draw first, what is you chance of winning if there is: a. replacement? This means that the game goes on indefinitely and the the probability that any one wins on a particular draw is 1/5. The probability that the first person to draw wins is: 1/5 + (4/5)^2 (1/5) + (4/5)^4 (1/5) + (4/5)^6 (1/5) + (4/5)^8 (1/5) + (4/5)^10 (1/5) + … = (1/5) [ 1 + (16/25)^1 + (16/25)^2 + (16/25)^3 + (16/25)^4 + (16/25)^5 + …] = (1/5) [ 1/(1-(16/25))] , since this is an infinite geometric series = (1/5) [1/(9/25)] = (1/5) (25/9) = 5/ b. no replacement? This means that there are just three ways for the first person to win. He wins on the first, third, or fifth draw. This problem is equivalent to 2.3-16a from last time. So, the probability of winning on a particular draw is always the same and is (1/5). The total probability that the first person wins is 3/. 2.5 – 2, 3, 6
Type A beans germinate 85% of the time and type B beans germinate 75% of the time. A bag of beans contains a mixture of these two types: 40% A and 60% B. a. Find the probability that a randomly selected seed will grow. P(G) = P(G ∩ A) U P(G ∩ B) = P(A) P(G | A) + P(B) P(G | B) = (0.4)(0.85) + (0.6)(0.75) = (0.34) + (0.45) = 0. b. Given that a seed grows, what is the probability that is type A? P(A | G) = P(A ∩ G) / P(G) = [P(A) P(G | A)] / P(G) = 0.34/ 0.79 = 0.
a. Find the probability of selecting a Belgian coin P(B20) = P(B20 ∩ C 1 ) U P(B20 ∩ C 2 ) = P(C 1 ) P(B20 | C 1 ) + P(C 2 ) P(B20 | C 2 ) = 0.25 (1/3) + 0.75 (1/2) = 11/

a. x 2 3 4 5 6 7 8 9 10 11 12 P(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/ f(x) = (6 - |x-7|)/36 , x = 2, 3, …, 12

a. f(w) = 1/12, x = 0,1, 2, …, 11

Probability Theory Homework Solutions: Independent Events, Combinatorics, and Dice Rolls, Assignments of Mathematics

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