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  • Three coins are tossed once. Find the probability of getting (iv) atmost 2 heads

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

(iv) at most \small 2 heads

Answers (1)

best_answer

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 heads = Event of getting 2 or fewer heads = {HHT, HTH, THH, TTH, HTT, THT}

\therefore n(E) = 6

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

The required probability of getting almost 2 heads is \frac{3}{4}.

Posted by

HARSH KANKARIA

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