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Question
A coin is tossed twice . what is the probability of getting
(ii) atmost 1 tail
Solution
When a coin is tossed twice , the outcomes are {HH , HT , TH , TT}
Number of outcomes = 4
Number of favourable outcomes = 3
P(getting atmost one tail) = `"Number of favourable outcomes"/"Total number of outcomes"` = `3/4`
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