a coin is tossed twice and the four possible

Question

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.