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1. In a simultaneous throw of two coins, the probability of getting at least one head is-

A. $\frac{1}{2}$

B. $\frac{1}{3}$

C. $\frac{2}{3}$

D. $\frac{3}{4}$

Solution:

Here S = {HH, HT, TH, TT}
Let E = event of getting at least one head = {HT, TH, HH}

∴ P (E) = \frac{n (E)}{n (S)} = \frac{3}{4}

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