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  • Three coins are tossed once. Find the probability of getting (vi) 3 tails

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

 (vi) \small 3 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 3 tails = {TTT}

\therefore n(E) = 1

\therefore P(E) = \frac{n(E)}{n(S)}  = \frac{1}{8} 

The required probability of getting 3 tails is \frac{1}{8}.

Posted by

HARSH KANKARIA

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