A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If two balls are…

Analyze the question: A box has 12 balls: 5 red, 4 blue, and the rest green (12 − 5 − 4 = 3 green). Two balls are drawn one after the other with replacement, and we want P(both red).

Concept: With replacement, the two draws are independent. The ball goes back into the box, so the probability stays the same on the second draw:

P(both red) = P(red) × P(red)

Solution:

• P(red on 1st draw) = 5/12
• P(red on 2nd draw) = 5/12 (the ball was replaced, so there are still 12 balls with 5 red)

P(both red) = 5/12 × 5/12 = 25/144

Check against the options: Option B, 5/33, is not the answer. It looks like a distractor, since the without-replacement answer would be 5/12 × 4/11 = 20/132 = 5/33. Because this question says "with replacement", we use 25/144.

Answer: A. 25/144

Tip: Always check the keywords "with replacement" or "without replacement."

• With replacement: the denominator stays the same (12 and 12).
• Without replacement: the denominator drops (12 and 11), and the number of favorable outcomes also drops.