Consider 3 urns. Urn A contains 2 white and 4 red balls,

Step 1 of 2

We have to find the probability of the ball from urn A was white given that exactly 2 white balls

were selected is \(P(E / F)=\frac{P(E \cap F)}{P(F)}\)

Let E be the event that the ball chosen from urn A is white

Let F be the event that exactly 2 white balls are selected

Given that urn A contains 2 white 4 red; total = 6

                  urn B contains 8 white 4 red; total = 12

                  urn C contains 1 white 3 red; total = 4

The total no.of ways of selecting a ball from each urn is

\(\left(\begin{array}{l}
6 \\
1
\end{array}\right)\left(\begin{array}{l}
12 \\
1
\end{array}\right)\left(\begin{array}{l}
4 \\
1
\end{array}\right)\)

=6(12)(4)

=288

No. of combinations with exactly 2 white balls 

Getting white balls from A, B and red ball from C;

then no.of combinations is \(\left(\begin{array}{l}
2 \\
1
\end{array}\right)\left(\begin{array}{l}
8 \\
1
\end{array}\right)\left(\begin{array}{l}
3 \\
1
\end{array}\right)=2(8)(3)=48\)

Getting white balls from A, C and red ball from B;

then no.of combinations is \(\left(\begin{array}{l}
2 \\
1
\end{array}\right)\left(\begin{array}{l}
4 \\
1
\end{array}\right)\left(\begin{array}{l}
1 \\
1
\end{array}\right)=2(4)(1)=8\)

Getting white balls from B, C and red ball from A;

then no.of combinations is \(\left(\begin{array}{l}
4 \\
1
\end{array}\right)\left(\begin{array}{l}
8 \\
1
\end{array}\right)\left(\begin{array}{l}
1 \\
1
\end{array}\right)=4(8)(1)=32\)

Then the no.of combinations with exactly 2 white balls is 48 + 8 + 32 = 88