If 3 coin is tossed simultaneously, then find the probability of getting at least two heads....
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answer is 1/2
s=[TTT,TTH,HTT,THT,HHH,HHT,HTH,THH]
n(S)=8
A=[HHH,HHT,HTH,THH]
n(A)=4
P(A)=n(A)/n(S)=4/8=1/2
s=[TTT,TTH,HTT,THT,HHH,HHT,HTH,THH]
n(S)=8
A=[HHH,HHT,HTH,THH]
n(A)=4
P(A)=n(A)/n(S)=4/8=1/2
Anu726:
thank you
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