A fair coin is tossed a fixed number of times. If the probability of getting 11 heads is equal to the probability of getting 13 heads, then the probability of getting 2 heads is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{69}{2^{22}}\)
Let the coin be tossed n times. The number of heads X has a binomial distribution with \(p = q = \frac{1}{2}\):
\( \displaystyle P(X = r) = {}^{n}C_{r}\left(\frac{1}{2}\right)^{n} \)
So P(X = 11) = P(X = 13) gives
\({}^{n}C_{11} = {}^{n}C_{13}\)
Since \({}^{n}C_{r} = {}^{n}C_{s}\) with \(r \ne s\) means r + s = n:
\(n = 11 + 13 = 24\)
Then
\( \displaystyle P(X = 2) = {}^{24}C_{2}\left(\frac{1}{2}\right)^{24} \)
\( \displaystyle = \frac{24 \times 23}{2} \times \frac{1}{2^{24}} = \frac{276}{2^{24}} \)
\( \displaystyle = \frac{4 \times 69}{2^{24}} = \frac{69}{2^{22}} \)