Three distinct coins are tossed together. Find the probability of getting

(i) at least 2 heads

(ii) at most 2 heads

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#### Solution

(i)The possible outcomes of the experiment are {HHH, HTH, HHT, HTT, THT, TTH, THH, TTT}.

So, the total number of possible outcomes = 8

The outcomes favourable to the event 'at least 2 heads' denoted by E, are {HHT, HTH, THH, HHH}

i.e. the number of outcomes favourable to E = 4

Hence, P(E) `=4/8=1/2`

(ii)The outcomes favourable to the event 'at most 2 heads' denoted by F, are {HHT, HTH, HTT, THT, TTH, THH, TTT}

i.e. the number of outcomes favourable to F = 7

Hence, P(F) = `7/8`

Concept: Basic Ideas of Probability

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