Question

A coin is tossed (m+n) times.Find the probability of at least m consecutive heads.

• A. $\frac{1+n}{2^{m+1}}$
• B. $\frac{2+n}{2^{m+1}}$
• C. $\frac{2+n}{2^m}$
• D. $\frac{1+n}{2^m}$

Answer 1

Answer: (B)

$\frac{2+n}{2^{m+1}}$

If the sequence of m consecutive heads starts from the first head i.e,$HHH...m$ times may be tail or head.
Probability of this event $=\frac{1}{2^m}$
Similarly if the sequence starts after the first tail, then its probability $=\frac{1}{2}\times \frac{1}{2^m}=\frac{1}{2^{m+1}}.$
If the sequence starts from ${(r+1)}^{th}$ throw, $r\ge 1,$ the first r throws may be heads or tails, but the $r^{th}$ throw must be a tail followed by m consecutive heads.
The prob.for this $=\frac{1}{2}\times \frac{1}{2^m}=\frac{1}{2^{m+1}}$
Hence reqd. prob.$=\frac{1}{2^m}+(\frac{1}{2^{m+1}}+\frac{1}{2^{m+1}}+\cdots ntimes)=\frac{2+n}{2^{m+1}}.$