in a simultaneous throw of two coins the probability of getting atleast one tail is
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1/4 is the answer.........
shammiasks:
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According to me
getting one tail
Let T= event of getting 1 tail. Then,T= {TH, HT} and, therefore, nT = 2.Therefore, P(getting 1 tail) = P(T) = n(T)/n(S) = 2/4 = 1/2
getting one tail
Let T= event of getting 1 tail. Then,T= {TH, HT} and, therefore, nT = 2.Therefore, P(getting 1 tail) = P(T) = n(T)/n(S) = 2/4 = 1/2
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