Question

A biased coin with probability $p$, (0$<p<$1) of showing head is tossed until a head appears for the first time. If the probability that the number of tosses required is odd is $\frac{7}{11}$ then $p$ =

• A. $\frac{4}{7}$
• B. $\frac{3}{7}$
• C. $\frac{4}{11}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{3}{7}$

One can get heads after 1 throw or 3 or 5 ... till infinity.
Let probability of a head = p and that of a tail = (1-p).
Hence, probability $=p+{(1-p)}^2\ast p+{(1-p)}^4\ast p+...\infty =\frac{p}{1-{(1-p)}^2}=\frac{1}{2-p}$
$\Rightarrow$, $\frac{1}{2-p}=\frac{7}{11}=>p=\frac{3}{7}$