Question : A coin is tossed 7 times. Then the probability that at least 4 consecutive heads appear is
(A) 3/16
(B) 5/32
(C) 5/16
(D) 1/8
Answer : option (B)
Solution :
Let H denote the head, T the tail and * any of the head or tail.
Then, P(H)=1/2 , P(T)=1/2 and P(*)=1. For at least four consecutive heads, we should have any of the following patterns:
| Probability | |
| (i) H H H H *** | \left( \dfrac{1}{2}\right) ^{4}\times 1=\dfrac{1}{16} |
| (ii) T H H H H ** | \left( \dfrac{1}{2}\right) ^{5}=\dfrac{1}{32} |
| (iii) * T H H H H * | \left( \dfrac{1}{2}\right) ^{5}=\dfrac{1}{32} |
| (iv) ** T H H H H | \left( \dfrac{1}{2}\right) ^{5}=\dfrac{1}{32} |
Since all the above cases are mutually exclusive, the probability of getting at least four consecutive heads (on adding) is
=1/16+3/32 = 5/32.