A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads, is

A
$15 / 2^{8}$
B
2/15
C
$15 / 2^{13}$
D
none of these

Solution:

Let n be the total number of tosses and X be the number of times head occurs. Then, X is a binomial variate with parameters n and p = 1/2.

Now,

$P (X = 7) = P (X = 9)$

$\begin{array}{l} \Rightarrow^{n} C_{7} {(\frac{1}{2})}^{7} {(\frac{1}{2})}^{n - 7} =^{n} C_{9} {(\frac{1}{2})}^{9} {(\frac{1}{2})}^{n - 9} \Rightarrow^{n} C_{7} =^{n} C_{9} \Rightarrow n = 16 \\ ∴ P (X = 2) =^{16} C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{16 - 2} = \frac{15}{2^{13}} \end{array}$

Solution:

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