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

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

  (ii) \small 2  heads

Answers (1)

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 2 heads = {HHT, HTH, THH}

\therefore n(E) = 3

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

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

Posted by

Ankit

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