Question

A coin is tossed twice . what is the probability of getting

(ii) atmost 1 tail

Answer 1

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`