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
$P\left(x\ge 5\right)={\left(\frac{3}{4}\right)}^4\frac{1}{4}+{\left(\frac{3}{4}\right)}^5\frac{1}{4}+{\left(\frac{3}{4}\right)}^6\left(\frac{1}{4}\right)+.\dots$
$=\frac{{\left(\frac{3}{4}\right)}^4\frac{1}{4}}{1-\frac{3}{4}}={\left(\frac{3}{4}\right)}^4$
$P\left(x\ge 3\right)={\left(\frac{3}{4}\right)}^2\frac{1}{4}+{\left(\frac{3}{4}\right)}^3\frac{1}{4}+{\left(\frac{3}{4}\right)}^4\left(\frac{1}{4}\right)+.\dots$
$=\frac{{\left(\frac{3}{4}\right)}^2\left(\frac{1}{4}\right)}{1-\frac{3}{4}}={\left(\frac{3}{4}\right)}^2$
$P\left(x\ge 5|x\ge 3\right)=\frac{P\left[\left(x\ge 5\right)\cap \left(x\ge 3\right)\right]}{P\left(x\ge 3\right)}=\frac{P\left(x\ge 5\right)}{P\left(x\ge 3\right)}=\frac{{\left(\frac{3}{4}\right)}^4}{{\left(\frac{3}{4}\right)}^2}=\frac{9}{16}$