find the probability of at least one head when two coins are tossed
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getting at least 1 head:
Let E6 = event of getting at least 1 head. Then, E6 = {HT, TH, HH} and, therefore, n(E6) = 3. Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾.
Let E6 = event of getting at least 1 head. Then, E6 = {HT, TH, HH} and, therefore, n(E6) = 3. Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾.
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