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

The correct answer is Option (1) → (A), (B) and (C) only

To determine the correct answer, we first calculate the total number of ways to draw 2 balls from the bag.

Total number of balls $= 4 \text{ (Red)} + 5 \text{ (Blue)} + 3 \text{ (Green)} = 12$

Total ways to draw 2 balls $= \begin{pmatrix}12\\2\end{pmatrix} = \frac{12 \times 11}{2 \times 1} = 66$

Step-by-Step Verification:

(A) Probability that both balls are red:

• Ways to pick 2 red: $\begin{pmatrix}4\\2\end{pmatrix} = \frac{4 \times 3}{2 \times 1} = 6$
• Probability $= \frac{6}{66} = \frac{1}{11}$
• Conclusion: Correct

(B) Probability that one 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 $= \frac{20}{66} = \frac{10}{33}$
• Conclusion: Correct

(C) Probability that both balls are blue:

• Ways to pick 2 blue: $\begin{pmatrix}5\\2\end{pmatrix} = \frac{5 \times 4}{2 \times 1} = 10$
• Probability $= \frac{10}{66} = \frac{5}{33}$
• Conclusion: Correct

(D) Probability that both balls are green:

• Ways to pick 2 green: $\begin{pmatrix}3\\2\end{pmatrix} = \frac{3 \times 2}{2 \times 1} = 3$
• Probability $= \frac{3}{66} = \frac{1}{22}$
• (Statement D claims 5/11, which is incorrect)
• Conclusion: Incorrect

Final Result:

Only statements (A), (B), and (C) are correct.