Question

Two coins look similar, but have different probabilities of falling "head". One is a fair coin, with $P(H)=1/2$, but the other is weighed so that $P(H)=4/5$. One of the coin is chosen at random and is tossed 10 times, let $X$ denotes the number of heads that of appear, and $F$ is the event that the fair coin was drawn. If $X=7$ is observed, the probability that the coin was fair, is

• A. $\frac{5^{10}}{5^8+8^{10}}$
• B. $\frac{5^8}{5^{10}+4^{10}}$
• C. $\frac{5^{10}}{5^{10}+8^7}$
• D. $\frac{5^{10}}{5^{10}+8^8}$

Answer 1

Answer: (D)

A : Coin randomly selected tossed 10 times, fell head wise 7 times
$B_1$ : coin was a fair coin $P\left(B_1\right)=1/2$
$B_2$ : Coin was a weighted coin $P\left(B_2\right)=1/2$
$P\left(A/B_1\right)={}^{10}{C}_7\cdot {\left(\frac{1}{2}\right)}^7\cdot {\left(\frac{1}{2}\right)}^3={}^{10}{C}_3\cdot \frac{1}{2^{10}}$
$P\left(A/B_2\right)={}^{10}{C}_7\cdot {\left(\frac{4}{5}\right)}^7\cdot {\left(\frac{1}{5}\right)}^3={}^{10}{C}_3\cdot \frac{4^7}{5^{10}}$
$P\left(B_1/A\right)=\frac{\frac{1}{2^{10}}}{\frac{1}{2^{10}}+\frac{4^7}{5^{10}}}=\frac{1}{1+\frac{4^7\cdot 2^{10}}{5^{10}}}=\frac{5^{10}}{5^{10}+8^8}$