**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.