Total probability and binomial distribution. This is taken from Ross book: when coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times. Compute the probability that the coin lands on heads on exactly 7 of the 10 flips.

jatericrimson8b

jatericrimson8b

Answered question

2022-09-16

Total probability and binomial distribution
This is taken from Ross book: when coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times. Compute the probability that the coin lands on heads on exactly 7 of the 10 flips.
What is incorrect behind this reasoning? let C be the event that the coin lands on heads. By total probability we have P ( C ) = 1 2 .7 + 1 2 .4 = .55. Then we apply a binomial distribution with n = 10, k = 7 and p = .55. However, this is incorrect. Why?

Answer & Explanation

Makhi Adams

Makhi Adams

Beginner2022-09-17Added 13 answers

Step 1
You don't have the event C, you have ten events C 1 , , C 10 where C j is the event of the j-th toss being heads. You are assuming these events are independent, but they are not.
Step 2
Essentially this is because it's the same coin involved in any pair of tosses. In detail the probability of C 1 C 2 is 1 2 ( 0.4 2 + 0.7 2 ) which isn't the square of P ( C i ) = 1 2 ( 0.4 + 0.7 )
Zackary Duffy

Zackary Duffy

Beginner2022-09-18Added 3 answers

Step 1
You have to condition first on the coin chosen, then calculate the probabilistic density function for each coin.
Let C be the event that coin 1 is chosen, and let X be the number of heads.
Step 2
P ( X = 7 ) = P ( X = 7 | C = 1 ) P ( C = 1 ) + P ( X = 7 | C = 2 ) P ( C = 2 ) = ( 10 7 ) ( 0.4 ) 7 ( 1 0.4 ) 3 1 2 + ( 10 7 ) ( 0.7 ) 7 ( 1 0.7 ) 3 1 2 = 3399 80038 + 2537 9508 = 5091 16460

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