A bag holds 3 red, 4 blue and 2 green counters. Two are taken at random without replacement. What is the probability that one is red and one is blue (in either order)?
Answer and explanation
Correct answer: A. 1/3
- Step 1: Red then blue: 3/9 x 4/8. Blue then red has the same value. Step 2. 2 x 3/9 x 4/8 = 1/3. Answer: 1/3. Why not the others: 1/6 counts only one order; 8/27 forgets that the bag has one fewer counter on the second draw; 17/54 uses the full bag for the blue-then-red draw.
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