I would like to know why the approach to the problem below is wrong: When coin A is flipped, it lands on heads with probability 0.4. When coin B is flipped, it lands on heads with probability 0.7. One of these coins is randomly chosen and flipped 10 times. Given that the first of these 10 flips lands heads, what is the conditional probability that exactly 7 of the 10 flips land on heads? Wrong approach: X: number of times the coin lands on heads H: first of these 10 flips lands heads A: coin A is chosen B: coin B is chosen P(X=7|H) = P(X=7, A|H) + P(X=7, B|H) = 9C6 x (0.4^6) x (0.6^3) x 0.5 + 9C6 x (0.7^6) x (0.3^3) x 0.5 = 0.1706 However, the correct answer is 0.1968.
I would like to know why the approach to the problem below is wrong: When coin A is flipped, it lands on heads with probability 0.4. When coin B is flipped, it lands on heads with probability 0.7. One of these coins is randomly chosen and flipped 10 times. Given that the first of these 10 flips lands heads, what is the conditional probability that exactly 7 of the 10 flips land on heads? Wrong approach: X: number of times the coin lands on heads H: first of these 10 flips lands heads A: coin A is chosen B: coin B is chosen P(X=7|H) = P(X=7, A|H) + P(X=7, B|H) = 9C6 x (0.4^6) x (0.6^3) x 0.5 + 9C6 x (0.7^6) x (0.3^3) x 0.5 = 0.1706 However, the correct answer is 0.1968.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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I would like to know why the approach to the problem below is wrong:
When coin A is flipped, it lands on heads with
Wrong approach:
X: number of times the coin lands on heads
H: first of these 10 flips lands heads
A: coin A is chosen
B: coin B is chosen
P(X=7|H) = P(X=7, A|H) + P(X=7, B|H) = 9C6 x (0.4^6) x (0.6^3) x 0.5 + 9C6 x (0.7^6) x (0.3^3) x 0.5 = 0.1706
However, the correct answer is 0.1968.
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