A coin is tossed 6 times. Find the probability of getting

(i) exactly 4 heads


(ii) at least 1 heads


(iii) at most 4 heads


(i) As the coin is tossed 6 times the total number of outcomes will be = 64


And we know that the favourable outcomes of getting exactly 4 heads will be 6= 15


Thus, the probability of getting exactly 4 heads will be



⇒ 15/64


(ii) As the coin is tossed 6 times the total number of outcomes will be = 64


And we know that the favourable outcomes of getting at least 1 heads will be 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 63


Thus, the probability of getting at least 1 head will be



⇒ 63/64


(iii) As the coin is tossed 6 times the total number of outcomes will be = 64


And we know that the favourable outcomes of getting at most 4 heads will be 6C0 + 6C1 + 6C2 + 6C3 + 6C4 = 57


Thus, the probability of getting at most 4 heads will be



⇒ 57/64


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