A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is:

We have two boxes and the first action is to pick one of them at random. Because no preference is mentioned, the probability of choosing either box is $$\frac12$$.

Let the event $$E$$ denote “exactly one white ball and one red ball are obtained in the two draws.” Let the event $$B$$ denote “the chosen box is box B.” Our aim is to find the conditional probability $$P(B\,|\,E)$$.

According to the definition of conditional probability,

$$P(B\,|\,E)=\dfrac{P(B\cap E)}{P(E)}.$$

We now compute the two probabilities in the fraction, first for “one white and one red” from each box, and then for the overall event.

Step 1: Probability of drawing one white and one red from box A.

Box A contains 2 white, 3 red and 2 black balls, a total of $$2+3+2 = 7$$ balls.

The number of ways of choosing any 2 balls from 7 is given by the combination formula $$^nC_r=\dfrac{n!}{r!\,(n-r)!},$$ so $$\binom{7}{2}.$$ Evaluating, $$\binom{7}{2}=\dfrac{7\cdot6}{2\cdot1}=21.$$

The favourable ways for exactly one white and one red are obtained by choosing 1 white out of 2 and 1 red out of 3, so $$\binom{2}{1}\binom{3}{1}=2\cdot3=6.$$

Thus,

$$P(E\text{ from }A)=\dfrac{6}{21}=\dfrac{2}{7}.$$

Step 2: Probability of drawing one white and one red from box B.

Box B contains 4 white, 2 red and 3 black balls, a total of $$4+2+3 = 9$$ balls.

The total number of pairs that can be chosen is $$\binom{9}{2}=\dfrac{9\cdot8}{2\cdot1}=36.$$

The favourable selections (1 white, 1 red) are counted by $$\binom{4}{1}\binom{2}{1}=4\cdot2=8.$$

Therefore,

$$P(E\text{ from }B)=\dfrac{8}{36}=\dfrac{2}{9}.$$

Step 3: Probability of the event $$E$$ overall.

Because each box is chosen with probability $$\frac12$$ and draws are made independently thereafter, the law of total probability gives

$$P(E)=P(A)\,P(E\text{ from }A)+P(B)\,P(E\text{ from }B).$$

Substituting $$P(A)=P(B)=\frac12$$ along with the results from Steps 1 and 2, we get

$$P(E)=\frac12\cdot\frac{2}{7}+\frac12\cdot\frac{2}{9}.$$

Combine the two fractions inside the parentheses first:

$$\frac{2}{7}+\frac{2}{9}=\frac{2\cdot9}{7\cdot9}+\frac{2\cdot7}{9\cdot7}=\frac{18+14}{63}=\frac{32}{63}.$$

Now multiply by $$\frac12$$:

$$P(E)=\frac12\cdot\frac{32}{63}=\frac{16}{63}.$$

Step 4: Joint probability $$P(B\cap E).$$

This is simply the product of choosing box B and then getting the desired colour combination:

$$P(B\cap E)=P(B)\,P(E\text{ from }B)=\frac12\cdot\frac{2}{9}=\frac{1}{9}.$$

Step 5: Required conditional probability.

Using the formula from the beginning,

$$P(B\,|\,E)=\dfrac{P(B\cap E)}{P(E)}=\dfrac{\frac{1}{9}}{\frac{16}{63}}.$$

To simplify, multiply numerator and denominator:

$$P(B\,|\,E)=\frac{1}{9}\times\frac{63}{16}=\frac{63}{144}.$$

We now reduce the fraction by dividing numerator and denominator by their greatest common divisor, which is 9:

$$\frac{63\div9}{144\div9}=\frac{7}{16}.$$

Hence, the correct answer is Option A.