1. A pair of fair dice is thrown. If the two numbers appearing are different, find the probability $p$ that: (a) the sum is 6; (b) an ace appears; (c) the sum is 4 or less. There are 36 ways the pair of dice can be thrown, and six of them, $(1, 1), (2, 2), \ldots, (6, 6)$, have the same numbers. Thus the reduced sample space will consist of $36 - 6 = 30$ elements. (a) The sum 6 can appear in four ways: $(1{,}5), (2, 4), (4, 2), (5, 1)$. (We cannot include $(3{,}3)$ since the numbers are the same.) Hence $p = \frac{4}{30} = \frac{2}{15}$. (b) An ace can appear in 10 ways: $(1{,}2), (1, 3), \ldots, (1, 6)$ and $(2, 1), (3, 1), \ldots, (6, 1)$. Therefore $p = \frac{10}{30} = \frac{1}{3}$. (c) The sum of 4 or less can occur in four ways: $(3, 1), (1, 3), (2, 1), (1, 2)$. Thus $p = \frac{4}{30} = \frac{2}{15}$. 7.22. A class has 12 boys and four girls. Suppose three students are selected at random from the class. Find the probability $p$ that they are all boys. The probability that the first student selected is a boy is $12/16$ since there are 12 boys out of 16 students. If the first student is a boy, then the probability that the second is a boy is $11/15$ since there are 11 boys left out of 15 students. Finally, if the first two students selected were boys, then the probability that the third student is a boy is $10/14$ since there are 10 boys left out of 14 students. Thus, by the multiplication theorem, the probability that all three are boys is $$p = \frac{12}{16} \cdot \frac{11}{15} \cdot \frac{10}{14} = \frac{11}{28}$$ (Probability, Dice, Sample Space, Ace, Contingency)

Answer: The probability that all three students selected are boys is $\frac{11}{28}$.

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