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

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

  (iii) atleast 2 heads

Answers (1)

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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 atleast 2 heads = Event of getting 2 or more heads = {HHH, HHT, HTH, THH}

\therefore n(E) = 4

\therefore P(E) = \frac{n(E)}{n(S)}  = \frac{4}{8} = \frac{1}{2} 

The required probability of getting atleast 2 heads is \frac{1}{2}.

Posted by

HARSH KANKARIA

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