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 denote the number of heads that of appears, and ‘F’ is the event that the fair coin was drawn. If X = 7 is observed, the probability that the coin was fair is

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$\frac{5^{10}}{5^{8} + 8^{10}}$

$\frac{5^{8}}{5^{10} + 4^{10}}$

$\frac{5^{10}}{5^{10} + 8^{7}}$

$\frac{5^{10}}{5^{10} + 8^{8}}$

answer is D.

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Detailed Solution

(4) Let A be the event, coin randomly selected tossed 10 times fell head wise 7 times
B1: coin was a fair coin
B2: coin is a weighed coin
$P (B_{1}) = \frac{1}{2} \begin{array}{l} P (B_{2}) = \frac{4}{5} \\ P (A / B_{1}) = 10 c_{7} {(\frac{1}{2})}^{7} {(\frac{1}{2})}^{3} \end{array}$
$P (A / B_{2}) = 10 c_{7} {(\frac{4}{5})}^{7} {(\frac{1}{5})}^{3} t h e n f i n d P (B_{1} / A)$