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  • Three coins are tossed once. Find the probability of getting (ix) atmost two tails

8. (ix) Three coins are tossed once. Find the probability of getting

 (ix) at most 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 at most 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT}

\therefore n(E) = 6

\therefore P(E) = \frac{n(E)}{n(S)}  = \frac{6}{8} = \frac{3}{4} 

The required probability of getting at most 2 tails is \frac{3}{4}.

Posted by

HARSH KANKARIA

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