Question

A biased coin is tossed repeatedly until a tail appears for the first time. The head is $3$ times likely to appear as the tail. The probability that the number of tosses required will be more than $4$ , given that in first two toss no tail has occur, is

• A. $\frac{3}{4}$
• B. $\frac{7}{8}$
• C. $\frac{9}{16}$
• D. $\frac{11}{32}$

Answer 1

Answer: (C)

$P\left(H\right)=\frac{3}{4},P\left(T\right)=\frac{1}{4}$
Let, $x$ be the number of tosses required
$P\left(x \geq 5\right)=\left(\frac{3}{4}\right)^{4}\frac{1}{4}+\left(\frac{3}{4}\right)^{5}\frac{1}{4}+\left(\frac{3}{4}\right)^{6}\left(\frac{1}{4}\right)+.\ldots $
$=\frac{\left(\frac{3}{4}\right)^{4} \frac{1}{4}}{1 - \frac{3}{4}}=\left(\frac{3}{4}\right)^{4}$
$P\left(x \geq 3\right)=\left(\frac{3}{4}\right)^{2}\frac{1}{4}+\left(\frac{3}{4}\right)^{3}\frac{1}{4}+\left(\frac{3}{4}\right)^{4}\left(\frac{1}{4}\right)+.\ldots $
$=\frac{\left(\frac{3}{4}\right)^{2} \left(\frac{1}{4}\right)}{1 - \frac{3}{4}}=\left(\frac{3}{4}\right)^{2}$
$P\left(x \geq 5 \left|\right. x \geq 3\right)=\frac{P \left[\left(x \geq 5\right) \cap \left(x \geq 3\right)\right]}{P \left(x \geq 3\right)}=\frac{P \left(x \geq 5\right)}{P \left(x \geq 3\right)}=\frac{\left(\frac{3}{4}\right)^{4}}{\left(\frac{3}{4}\right)^{2}}=\frac{9}{16}$