A fair coin is tossed a fixed number of times.

Question

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :

Accepted Answer

Let the coin be tossed n-times

$$P(H) = P(T) = {1 \over 2}$$

P(7 heads) = $${}^n{C_7}{\left( {{1 \over 2}} \right)^{n - 7}}{\left( {{1 \over 2}} \right)^7} = {{{}^n{C_7}} \over {{2^n}}}$$

P(9 heads) = $${}^n{C_9}{\left( {{1 \over 2}} \right)^{n - 9}}{\left( {{1 \over 2}} \right)^9} = {{{}^n{C_9}} \over {{2^n}}}$$

P(7 heads) = P(9 heads)

$${}^n{C_7} = {}^n{C_9} \Rightarrow n = 16$$

P(2 heads) = $${}^{16}{C_2}{\left( {{1 \over 2}} \right)^{14}}{\left( {{1 \over 2}} \right)^2} = {{15 \times 8} \over {{2^{16}}}}$$

P(2 heads) $$ = {{15} \over {{2^{13}}}}$$

Solutions · 1