Question

A coin is tossed m + n times, where $m\ge n.$ The probability of getting at least m consecutive heads is

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is B

Here $P\left(H\right)=P\left(T\right)=\frac{1}{2}$ and $P\left(X\right)=1,$ where X denotes head or tail.
If the sequence of m consecutive heads starts from the first throw, we have
(HH ….. m times) (XX …. n times)
$\therefore$ Chance of this event $=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}......$ m times $=\frac{1}{2^m}$
$\because$ m + 1 and subsequent throws may be head or tail since we are considering at least m consecutive heads. If the sequence of m consecutive heads starts from the second throw, the first must be a tail and we have, the chance of this event $=\frac{1}{2}.\frac{1}{2^m}=\frac{1}{2^{m+1}}$
If the sequence of heads starts from $(r+1{)}^{th}$ throw the first (r – 1) throws may be head or tail but $r^{th}$ throw must be a tail and we have,
(XX ……… (r – 1)times) T (HH ……… m times)
(XX ……. n – m – r times)
The chance of this event also $\frac{1}{2}\times \frac{1}{2^m}=\frac{1}{2^{m+1}}$
Since all the above events are mutually exclusive, so the required probability
$=\frac{1}{2^m}+(\frac{1}{2^{m+1}}+\frac{1}{2^{m+1}}+....ntimes)$
$=\frac{1}{2^m}+\frac{n}{2^{m+1}}=\frac{n+2}{2^{m+1}}$
Note : Students should remember this question as a formula.