Question

Consider the experiment of throwing a die, if a multiple of 3 comes up throw the die again and if any other number comes toss a coin. Find the conditional probability of the event “the coin shows a tail”, given that “at least one die shows a 3''.

Answer 1

S={(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,H), (2,H), (3,H), (4,H), (5,H), (1,T), (2,T), (3,T), (4,T), (5,T)}
∴n(S)=20

P (first die shows a multiple of 3) = \(\frac {12}{36}\) = \(\frac 13\)

P (first die shows a number which is not a multiple of 3) = \(\frac 46×\frac 12+\frac 46×\frac 12\) = \(\frac {8}{12}\) = \(\frac 23\)

Let,
A = the coin shows a tail = {(1, T), (2, T), (4, T), (5, T)}
B = at least one die shows a 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6,3)}

\(A∩B = ϕ\)
\(n(A) = 4\)
\(n(B) = 7\)
\(n(A∩B) = 0\)

\(P(B) =\) \(\frac {6}{36}\) =\(\frac 16\) 

\(and,\  P(A∩B)=0\)

\(P(A|B)=\frac {P(A∩B)}{P(B)}\)

\(P(A|B)=\frac {0}{\frac {7}{36}}\)

\(P(A|B)=0\)