A coin is tossed twice and the four possible outcomes are assumed to be equally likely. If A is the event, both head and tail have appeared , and B be the event at most one tail is observed find
P(A/B)
A) 1/3 B) 2/3 C) ½ D) ¼
Here, S={HH,HT,TH,TT},A={HT,TH} and B={HH,HT,TH,}
∴A∩B={HT,TH}
Now(A)=n(A)/n(s)=2/4=1/2,P(B)=n(B)/n(S)=3/4
And (A∩B)=n(A∩B)/n(S)=2/4=1/2
∴P(A/B)=P(A∩B)/P(B)=1/2/3/4 = 2/3.