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
Answer & explanation
Correct answer: option 3
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 = $\frac{1}{2}.$
Now,
$P(X=7)=P(X=9)$
$⇒{^nC}_7 \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{n-7}={^nC}_9 \left(\frac{1}{2}\right)^9 \left(\frac{1}{2}\right)^{n-9}⇒{^nC}_7={^nC}_9⇒n = 16 $
$∴P(X=2)= {^{16}C}_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{16-2}=\frac{15}{2^{13}}$