I know how to do the first part. I’m not sure about the second and third part. C
ID: 3049543 • Letter: I
Question
I know how to do the first part. I’m not sure about the second and third part.Consider the following game. You roll a fair die. If you roll a 5 or 6 you select a ball from Jar 2. Otherwise, you select a ball from Jar 1. Jar 1 contains 15 red balls and 5 green balls while Jar2 contains 8 red balls and 12 green balls. You will play the game once Problem 1. What is the probability that you select a red ball? (10 points) Problem 2. Find the Probability of (Roll 5 or 6 given Red Ball)? In other words, given that you select a Red ball, what is the probability you would roll a 5 or 6? (10 points) Problem 3. Find Probability (Roll 1 through 4 Union Red Ball). (8 points) Problem 4. Find Probability (Roll 1 through 4 Intersection Green Ball). 7 points) There is an additional aspect to the game that we have not yet mentioned. You are given $1 if you roll a 5 or 6 and S3 if you roll a 1 through 4. Let the random variable X represent your profit from the die part of the game. Ifyou select a red ball, you are given S4. If you select a green ball, you are given S2. Let the random variable Y represent the profit from the ball selection part of the game. Let Z = X + Y. You will play the game once Problem 4. Construct the probability distribution function for your profit from the entire game. (12 points) Problem 5. Find E(Z) and g. (11 points) Problem 6. Find E(XY). (11 points) Problem 7. Find ..y (in other words, find the correlation of X and Y). (11 points) Assume that one would play the above game ten times and sampling is done with replacement. Problem 8. What is the Probability that you would select a green ball exactly seven times (out of the ten games)? (10 points) Problem 9. What is the probability that your profit would be S7 exactly three times (out of the ten games)? (10 points)
Explanation / Answer
Let A represent the event that roll of the die gives 5 or 6 up and so P(A) = 2/6 = 1/3 …….(1)
Then, trivially,
AC represents the event that roll of the die gives 1, 2, 3 or 4 up and so P(AC) = 4/6 = 2/3 ...(2)
Let B represent the event that a red ball is picked. Then,
if A, P(B) = P(Red ball from Jar 2) = 8/20 [Jar 2 has 20 balls of which 8 are red] …….(3)
if A, P(BC) = P(Green ball from Jar 2) = 12/20 [Jar 2 has 20 balls of which 12 are green] (3A)
if AC, P(B) = P(Red ball from Jar 1) = 15/20 [Jar 1 has 20 balls of which 15 are red] …….(4)
if AC, P(BC) = P(Green ball from Jar 1) = 5/20 [Jar 1 has 20 balls of which 5 are green]…(4a)
Back-up Theory
If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(5)
P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(6)
P(A/B) = P(B/A) x { P(A)/P(B)}……………………………..………………….(7)
Pr 1
Probability of selecting a red ball = P(B)
= {P(B/A) x P(A)} + {P(B/AC) x P(AC)} [vide (6) above]
= {(8/20) x (1/3)} + {(15/20) x (2/3)
= 19/30 ANSWER
Pr 2
Probability of roll 5 or 6 given red ball is picked
= P(A/B)
= P(B/A) x P(A)/P(B) [vide (7) above]
= {(8/20) x (1/3)}/(19/30) [19/30 from Answer of Pr 1]
= 4/19 ANSWER
Pr 3
Probability of roll 1, 2, 3 or 4 given red ball is picked
= P(AC/B)
= P(B/AC) x P(AC)/P(B) [vide (7) above]
= {(15/20) x (2/3)}/(19/30) [19/30 from Answer of Pr 1]
= 15/19 ANSWER
Pr 4
Probability of roll 1, 2, 3 or 4 intersection green ball
= (Probability of roll 1, 2, 3) x (Probability of green ball from Jar 1)
= (2/3) x (5/20)
= 1/6 ANSWER