Question

A normal coin is continued tossing unless a head is obtained for the first time. If the probability that number of tosses needed are at most 3 is p and the probability that number of tosses are even is q, then find the value of 24 pq.

Answer 1

First we need to find the probability that number of tosses are at most 3

If we got a head on first probability

Head on second toss $=\frac{1}{2}\times \frac{1}{2}$

Head on third toss $=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}$

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

In similar way,

$q=\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+......=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{3}{4}$

So, 24pq=7.