Problem c.) There are three possible ways to end up with exactly 1 woman in three selections: (M **and** M **and** W) **or** (M **and** W **and** M) **or** (W **and** M **and** M). The associated probabilities are P(M&M&W) = (8/10)×(7/9)×(2/8) = 14/90, P(M&W&M) = (8/10)×(2/9)×(7/8) = 14/90 and P(W&M&M) = (2/10)×(8/9)×(7/8) = 14/90. Any of these three outcomes will get the desired end result, and so the probability of ending up with exactly 1 woman is P(1W) = (14/90)+ (14/90)+ (14/90) = 42/90 = 7/15.

Problem d.) There are three possible ways to end up with exactly 2 women in three selections: (M **and** W **and** W) **or** (W **and** M **and** W) **or** (W **and** W **and** M). The associated probabilities are P(M&W&W) = (8/10)×(2/9)×(1/8) = 1/45, P(W&M&W) = (2/10)×(8/9)×(1/8) = 1/45 and P(W&W&M) = (2/10)×(1/9)×(8/8) = 1/45. Any of these three outcomes will get the desired end result, and so the probability of ending up with 2 women is P(2W) = (1/45)+ (1/45)+ (1/45) = 3/45 = 1/15.

Problem e.) There are two ways to end up with at least one woman: 1W **or** 2W. The associated probabilities are P(1W) = 7/15 and P(2W) = 1/15. Any of these two outcomes will get the desired end result, and so the probability of ending up with at least 1 woman is P(≥1W) = 7/15+1/15 = 8/15.

The above methods show the value of *understanding* probability problems in terms of first principles (pigeonholing & counting), as well as the mathematical meaning of “and” and “or” when calculating the probabilities.