A bag holds 4 red, 7 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: B. 14/39
- Step 1: Red then blue: 4/13 x 7/12. Blue then red has the same value. Step 2. 2 x 4/13 x 7/12 = 14/39. Answer: 14/39. Why not the others: 7/39 counts only one order; 56/169 forgets that the bag has one fewer counter on the second draw; 175/507 uses the full bag for the blue-then-red draw.
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