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Three unbiased coins are tossed. What is the probability of getting at least 2 heads ?

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\frac{1}{4}

$\frac{1}{8}$

\frac{1}{3}

$\frac{1}{2}$

answer is D.

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Detailed Solution

Here

S = {T T T, T T H, T H T, H T T, T H H, H T H, H H T, H H H}

Let $E =$ even of getting at least two heads $= {T H H, H T H, H H T, H H H}$

$P (E) = \frac{n (E)}{n (S)} = \frac{4}{8} = \frac{1}{2}$

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