Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

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

Options:

$\frac{15}{2^8}$

$\frac{2}{15}$

$\frac{15}{2^{13}}$

none of these

Correct Answer:

$\frac{15}{2^{13}}$

Explanation:

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 $

$∴P(X=2)= {^{16}C}_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{16-2}=\frac{15}{2^{13}}$