Binomial distribution and conditional probability

[Yankel]

Hello all. I saw this problem in a book. I tried solving it, and compared it to the suggested solution. Results don't match, and I think that I am correct. Could you please help me decide what the right answer is ?

This is the question:

When coin 1 is flipped, it lands on heads with probability 0.4; when coin 2 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 ten flips lands heads, what is the conditional probability that exactly 7 of the 10 flips land on heads?

My final answer is:

\[\frac{0.5\cdot 0.4\cdot \binom{9}{6}\cdot 0.4^{6}\cdot 0.6^{3}+0.5\cdot 0.7\cdot \binom{9}{6}\cdot 0.7^{6}\cdot 0.3^{3}}{0.5\cdot 0.4+0.5\cdot 0.7}\]

while the book's answer is almost the same, only that the second

\[\binom{9}{6}\]

from my solution is missing. Maybe a typo mistake ?

Thank you.

 

[jvicens]

Hello there,

Thank you for bringing this discrepancy to our attention. it is important to thoroughly check and verify our solutions before drawing any conclusions. After reviewing your calculations and comparing them to the suggested solution, I believe you are indeed correct and the book's answer may have a typo mistake.

The second binomial coefficient, $\binom{9}{6}$, is necessary in the calculation because it represents the number of ways to choose 6 out of 9 flips to land on heads, given that the first flip has already landed on heads. This is a crucial step in the calculation of the conditional probability.

I would recommend reaching out to the author or publisher of the book to inform them of this error. It is important for future readers to have accurate solutions to refer to. Thank you for bringing this to our attention and for being diligent in your problem-solving approach. Keep up the good work!