**8.1** . (In fact, we *always* get doubles with probability when at least one of the dice is fair.) Any two faces whose sum is 7 have the same probability in distribution Pr_{1}, so S = 7 has the same probability as doubles.

**8.2** There are 12 ways to specify the top and bottom cards and 50! ways to arrange the others; so the probability is 12.50!/52! = 12/(51.52) = .

**8.3** (3 + 2 + · · · + 9 + 2) = 4.8; (3^{2} + 2^{2} + · · · + 9^{2} + 2^{2} – 10(4.8)^{2}) = , which is approximately ...

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