A box contains 6 red balls, 4 blue balls, and 2 green balls. If two balls are drawn randomly without replacement, what is the probability that both balls are red?
- A1/11
- B15/44
- C3/11
- D5/22
Show answer & explanationAnswer & explanation
Correct answer: D. 5/22
Total number of balls initially = 6 (red) + 4 (blue) + 2 (green) = 12 balls. Probability of drawing the first red ball: P(1st Red) = Number of red balls / Total balls = 6/12 = 1/2. Since the first ball is not replaced, there are now 11 balls left, and 5 of them are red. Probability of drawing the second red ball (given the first was red): P(2nd Red | 1st Red) = 5/11. The probability that both balls are red is the product of these probabilities: P(Both Red) = P(1st Red) * P(2nd Red | 1st Red) = (6/12) * (5/11) = (1/2) * (5/11) = 5/22.
Why the other options are wrong
- A. Incorrect. This might arise from errors in calculating either probability or their product.
- B. Incorrect. A common error like (6/12) * (6/11) = 3/11, but 15/44 is a different error.
- C. Incorrect. Possibly from (6/12) + (5/11) or other miscalculation.
Conditional Probability (Without Replacement)
Conditional probability without replacement involves calculating the probability of a sequence of events where the outcome of the first event changes the sample space (and thus the probabilities) for subsequent events. The key is that items are not returned to the pool.
- The denominator (total possible outcomes) decreases with each draw.
- The numerator (favorable outcomes) also decreases if an item of that type is drawn.
- The probability of multiple events occurring in sequence is the product of their individual (conditional) probabilities.
Memory trick: Draw, then discard; the next chance is changed by the first card.