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Question
Three distinct coins are tossed together. Find the probability of getting
(i) at least 2 heads
(ii) at most 2 heads
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`
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