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Three coins are tossed once. Find the probability of getting (vii) exactly two tails

Q : 8     Three coins are tossed once. Find the probability of getting

             (vii) exactly two tails

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Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}

\therefore n(E) = 3

\therefore P(E) = \frac{n(E)}{n(S)}  = \frac{3}{8} 

The required probability of getting exactly 2 tails is \frac{3}{8}.

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