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 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

1. A

$15/{2}^{8}$

2. B

2/15

3. C

$15/{2}^{13}$

4. D

none of these

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### 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,

$P\left(X=7\right)=P\left(X=9\right)$

$\begin{array}{l}{⇒}^{n}{C}_{7}{\left(\frac{1}{2}\right)}^{7}{\left(\frac{1}{2}\right)}^{n-7}{=}^{n}{C}_{9}{\left(\frac{1}{2}\right)}^{9}{\left(\frac{1}{2}\right)}^{n-9}{⇒}^{n}{C}_{7}{=}^{n}{C}_{9}⇒n=16\\ \therefore P\left(X=2\right){=}^{16}{C}_{2}{\left(\frac{1}{2}\right)}^{2}{\left(\frac{1}{2}\right)}^{16-2}=\frac{15}{{2}^{13}}\end{array}$

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