Question

A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable Y denote the number of heads. The value of var(Y). where var(.) denotes the variance, equals:

• A. $\frac{7}{8}$
• B. $\frac{49}{64}$
• C. $\frac{7}{64}$
• D. $\frac{105}{64}$

Answer 1

The correct option is C $\frac{7}{64}$

Given the maximum no. of tosses = 3


y = {No. of Heads} = {1, 0}

P(y = 1H) = P(H) or P(TH) or P(TTH)

$=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}$

$P(y=0H)=P\left(TTT\right)=\frac{1}{8}$

So, probability distribution is:

$\begin{array}{ccc}y:& 0& 0\\ P\left(y\right):& \frac{1}{8}& \frac{7}{8}\end{array}$

$\left(Y\right)=y_{1}P\left(y_1\right)+y_{2}P\left(y_2\right)$

$=0\times \frac{1}{8}+1\times \frac{7}{8}=\frac{7}{8}$

$E\left(Y^2\right)=\times 0^2\times \left(\frac{1}{7}\right)+1^2\times \left(\frac{7}{8}\right)$

$=\frac{7}{8}$

$\therefore \text{Variance (Y)}=E\left(Y^2\right)-\left[E\right(Y){]}^2$

$=\frac{7}{8}-{\left(\frac{7}{8}\right)}^2$

$=\frac{7}{8}-\frac{49}{64}$

$=\frac{56-49}{64}=\frac{7}{64}$