A bag contains 7 red and 4 black identical balls. Two balls were picked at random from the bag and replaced…

Analyze the question: A bag has 7 red and 4 black balls, 11 in total. Two balls are picked, and each ball is replaced after it is picked. We want the probability that both balls are the same colour.

Concept: Because the balls are replaced, the two picks are independent. The probabilities stay the same on each draw:

• P(red) = 7/11
• P(black) = 4/11

Solution:

P(both red) = 7/11 × 7/11 = 49/121

P(both black) = 4/11 × 4/11 = 16/121

P(same colour) = 49/121 + 16/121 = 65/121

Check: Without replacement, the answer would be (7×6 + 4×3)/(11×10) = 54/110 = 27/55. That is option D, a distractor for students who ignore the word "replaced." The question says the balls were replaced, so the correct working is the one above.

Answer: Option B, 65/121.

Tip: Look for the words "with replacement" or "without replacement."

• With replacement: the denominators stay the same (11 × 11), and the draws are independent.
• Without replacement: the total drops by 1 on the second draw (11 × 10).

For "same colour," always add the separate cases (both red + both black).