A bag contains 4 red, 5 blue and 3 green balls. If two balls are draw…

The correct answer is Option (2) → (A), (B) and (C) only

1. Total Number of Outcomes

The bag contains:

• Red (R) = 4
• Blue (B) = 5
• Green (G) = 3
• Total (N) = $4 + 5 + 3 = 12$

Total ways to choose 2 balls from 12:

$^nC_r = \begin{pmatrix}12\\2\end{pmatrix} = \frac{12 \times 11}{2 \times 1} = 66$

2. Evaluating the Statements

(A) Probability that both balls are red:

• Ways to pick 2 red balls from 4: $\begin{pmatrix}4\\2\end{pmatrix} = \frac{4 \times 3}{2} = 6$
• Probability $P(RR) = \frac{6}{66} = \frac{1}{11}$
• Result: Correct.

(B) Probability that one ball is red and one is blue:

• Ways to pick 1 red and 1 blue: $\begin{pmatrix}4\\1\end{pmatrix} \times \begin{pmatrix}5\\1\end{pmatrix} = 4 \times 5 = 20$
• Probability $P(RB) = \frac{20}{66} = \frac{10}{33}$
• Result: Correct.

(C) Probability that both balls are blue:

• Ways to pick 2 blue balls from 5: $\begin{pmatrix}5\\2\end{pmatrix} = \frac{5 \times 4}{2} = 10$
• Probability $P(BB) = \frac{10}{66} = \frac{5}{33}$
• Result: Correct.

(D) Probability that both balls are green:

• Ways to pick 2 green balls from 3: $\begin{pmatrix}3\\2\end{pmatrix} = 3$
• Probability $P(GG) = \frac{3}{66} = \frac{1}{22}$
• Statement says $5/11$ (which is $30/66$).
• Result: Incorrect.

Conclusion

Statements (A), (B), and (C) are correct.