Question

A biased coin with probability $p,0<p<1,$ of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is $2/5,$ then $p$ equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{2}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{1}{3}$

The probability of first head occurring at even tosses=Probability of head occurring at second toss(i.e $\left(1-p\right)\times p)+$ probability of head occurring at fourth toss (i.e ${\left(1-p\right)}^3\times p)+$probability of head (i.e ${\left(1-p\right)}^3\times p)$ occouring at sixth toss (i.e ${\left(1-p\right)}^5\times p)+.......$

$i.e\left(\frac{2}{5}\right)=\left(1-p\right)\times p\left[{\left(1-p\right)}^2+{\left(1-p\right)}^4+.....\right]$

Solve this using sum of infinite $GP$ to get $p=\left(\frac{1}{3}\right)$