Answer: (v) getting no tail: Let E5 = event of getting no tail. Then, E5 = {HH} and, therefore, n(E5) = 1. Therefore, P(getting no tail) = P(E5) = n(E5)/n(S) = ¼. Reply
Answer:
(v) getting no tail:
Let E5 = event of getting no tail. Then, E5 = {HH} and, therefore, n(E5) = 1. Therefore, P(getting no tail) = P(E5) = n(E5)/n(S) = ¼.
Answer:
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