Question

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to that getting 9 heads, then the probability of getting 3 heads is

• A. $\frac{35}{2^{12}}$
• B. $\frac{35}{2^{14}}$
• C. $\frac{7}{2^{12}}$
• D. $\frac{7}{2^{14}}$

Answer 1

The correct option is A $\frac{35}{2^{12}}$

Probability of r successes in n trials is equal to ${}^n{C}_r{p}^r{q}^{n-r}$, where p is the probability of success in one trial and q is the probability of failure in one trial.
Let the coin be tossed n times. Here p and q are both 1/2
$P\left(getting7heads\right)=n{C}_7{\left(\frac{1}{2}\right)}^{n}andP\left(getting9heads\right)=n{C}_9{\left(\frac{1}{2}\right)}^{n}Given,P\left(7heads\right)=P\left(9heads\right)\Rightarrow n{C}_7=n{C}_9\Rightarrow n=16\therefore P\left(3heads\right)=16C_3{\left(\frac{1}{2}\right)}^{16}=\frac{35}{2^{12}}.$