Question

A coin is tossed $3$ times by $2$ persons. What is the probability that both get an equal number of heads

• A. $\frac{3}{8}$
• B. $\frac{1}{9}$
• C. $\frac{5}{16}$
• D. None of these

Answer 1

The correct option is C

$\frac{5}{16}$

The explanation for the correct option:

Step 1: First find the probability of success(getting head) and failure(getting tail):

We have been given that a coin is tossed three times by two persons.

We need to find the probability that both gets an equal number of heads.

Let us assume that $A,B$ are two persons.

Let, event $M$: number of heads obtained by person $A$

And event $N$: number of heads obtained by person $B$

For both the events $M,N$,

$p=\frac{1}{2},q=\frac{1}{2},n=3$

Step 2: Find the probability of each event:

Therefore, the probability that both $A$ and $B$ will get an equal number of heads is,

$=P(M=0)P(N=0)+P(M=1)P(N=1)+P(M=2)P(N=2)+P(M=3)P(N=3)$

$=\left[\left({}^{3}C_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{3-0}\right)\times \left({}^{3}C_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{3-0}\right)\right]+\left[\right({}^{3}C_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{3-1})\times ({}^{3}C_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{3-1}\left)\right]+\left[\left({}^{3}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{3-2}\right)\times \left({}^{3}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{3-2}\right)\right]+\left[\left({}^{3}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{3-3}\right)\times \left({}^{3}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{3-3}\right)\right]$

$={\left[{}^{3}C_0{\left(\frac{1}{2}\right)}^3\right]}^2+{\left[{}^{3}C_1\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^2\right]}^2+{\left[{}^{3}C_2{\left(\frac{1}{2}\right)}^2\left(\frac{1}{2}\right)\right]}^2+{\left[{}^{3}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^0\right]}^2$

$$\begin{gathered} ={\left(\frac{3!}{0!(3-0)!}\times \frac{1}{8}\right)}^2+{\left(\frac{3!}{1!(3-1)!}\times \frac{1}{8}\right)}^2+{\left(\frac{3!}{2!(3-2)!}\times \frac{1}{8}\right)}^2+{\left(\frac{3!}{3!(3-3)!}\frac{1}{8}\right)}^2 \\ =\left(\frac{3!}{3!}+\frac{3!}{2!}+\frac{3!}{2!}+\frac{3!}{3!}\right)\times {\left(\frac{1}{8}\right)}^2 \\ =(1+9+9+1)\times {\left(\frac{1}{8}\right)}^2 \\ =20\times \frac{1}{64} \\ =\frac{5}{16} \end{gathered}$$

Hence, the correct option is (C).