Question

Three boxes contain respectively 3 whites and 1 black, 2 white and 2 black, 1 white and 3 black balls, from each of the boxes one ball is drawn at random. The probability that 2 white and 1 black balls will be drawn is

A. \[\dfrac{{13}}{{32}}\]

B. \[\dfrac{1}{4}\]

C. \[\dfrac{1}{{32}}\]

D. \[\dfrac{3}{{16}}\]

Answer 1

Hint: First see the total number of balls that are to be drawn which is 3. Out of these 3 see how many are white and how many are black. Figure out the combination these balls can make such that you can analyze the possible scenarios in which they can be drawn and hence the final probability.

Complete Step-by-Step solution:

There are 3 ways to draw 2 white and 1 black ball from three boxes. They are:

Let 1,2,3 be the sequence of these boxes, W be the white balls, B be the black balls

So, First way is W,W,B

Second way is W,B,W          

Third way is B,W,W

Now we will see the probability

First box: P (white ball) = \[\dfrac{3}{4} = 0.75\] (\[\because \]total no. of balls in box 1 is 4, out of which 3 are white balls)

∴ P (black ball) = (1- P (white ball)) = 0.25

Similarly, Second box: P (white balls) =\[\dfrac{2}{4}\]= 0.5

∴ P (black ball) = 0.5

Similarly, Third box: P (white ball) = \[\dfrac{1}{4}\]= 0.25

∴ P (black ball) = 0.75

Now we will calculate the probability of the three ways mention in the starting of the solution

The P (first way) = probability of getting a white ball from first box × probability of getting a white from second box × probability of getting a black ball from third box

i.e. 0.75 × 0.5 × 0.75

= 0.28125

Similarly, P (second way) = 0.75 × 0.5 × 0.25

= 0.09375

Similarly, P (third way) = 0.25 × 0.5 × 0.25

= 0.03125

Thus the probability of getting 2 white and 1 black ball=P (first way)+P (second way)+P (third way) = 0.28125+0.09375+0.03125

= 0.40625

= \[\dfrac{{13}}{{32}}\]

∴ The correct option is ‘(A)’.

Note: In this question the three events are of independent nature, the general formula for their probability is given by the formula below, say these three events are A, B, C

∴\[P(A \cap B \cap C) = P\left( A \right).P\left( B \right).P\left( C \right)\]