Two similar boxes B_i (i = 1, 2) contain (i + 1) red and (5 – i – 1) black balls. One box is chosen at random and two balls are drawn randomly. What is the probability that both the balls are of different colours?

Concept:

• \(n_{C_{r}}\) → choosing or selecting r elements from n elements.
• \(n_{C_{r}}=\frac{n !}{(n-r) !(r) !}\)
• \(n_{C_{1}} = n\)

Calculation:

Clearly, B1​ has 2 red and 3 black balls whereas B2​ has 3 red and 2 black balls.

P(B1) = P(B2) = 1/2 

We need probability of getting both the balls of different colours

i.e., Required probability, P = P(B1)P(B)P(R) + P(B2)P(B)P(R)

Here for bag 1, we need to select 1 black and 1 red out of 3 black and 2 red balls respectively 

Similarly, for bag 2

P = P(B1)P(B)P(R) + P(B2)P(B)P(R) 

⇒ \(\frac{1}{2} × \frac{{ }^{3} C_{1} ×{ }^{2} C_{1}}{{ }^{5} C_{2}}+\frac{1}{2} × \frac{{ }^{3} C_{1} × { }^{2} C_{1}}{{ }^{5} C_{2}}\) 

(∵ Choosing one black and red out of 3 and 2 respectively from the B₁ of 5 balls

Similarly, from B₂ of 5 balls choosing one black and red out of 2 and 3 respectively.)

 \(⇒ \frac{1}{2} × \frac{2 × 3}{\frac{5 × 4}{2}}+\frac{1}{2} × \frac{3 × 2}{\frac{5 × 4}{2}}\)    (∵​\(5_{C_{2}} = \frac{5\times 4}{2\times 1}\) and \(n_{C_{1}} = n\))​

⇒ 3/10 + 3/10

⇒ 6/10

⇒ 3/5

Hence, option (4) is correct.​