A bag Contains 2 red, 3 blue and 4 green identical balls. If two balls are picked at random one after the…

Analyze the question: We have 9 balls in total (2 red + 3 blue + 4 green). Two balls are drawn one after the other without replacement, and we want the probability that both are green.

Concept: For draws without replacement, the total number of balls and the number of green balls both drop after the first draw. We multiply the probabilities of the two events:

P(both green) = P(1st green) × P(2nd green | 1st green)

Solution:

• Total balls = 2 + 3 + 4 = 9
• P(1st green) = 4/9
• After one green ball is removed, 8 balls remain, 3 of them green.
• P(2nd green) = 3/8

P(both green) = 4/9 × 3/8 = 12/72 = 1/6

Check with combinations: P = ⁴C₂ / ⁹C₂ = 6/36 = 1/6 ✓

Answer: The result is 1/6, which is option A.

Tip: In "without replacement" questions, reduce both the numerator and the denominator by 1 for the second draw. Always simplify at the end. Here 12/72 reduces to 1/6. Options like 16/18 are traps, since a probability that large is impossible for this situation.