chapter 1

Chapter 2: ProbabilitySection 2.329.a. (5)(4) = 20 (5 choices for president, 4 remain for vice president)b. (5)(4)(3) = 60⎛5⎞⎝2⎠5!2!3!c. ⎜ ⎟ = = 10 (No ordering is implied in the choice)30.a. Because order is important, we’ll use P 8,3 = 8(7)(6) = 336.b. Order doesn’t matter here, so we use C 30,6 = 593,775.⎛8⎞⎝2⎠⎛10⎞⎝ 2 ⎠⎛12⎞⎝ 2 ⎠c. From each group we choose 2: ⎜ ⎟ • ⎜ ⎟ • ⎜ ⎟ = 83, 16083,160593,775d. The numerator comes from part c and the denominator from part b: = . 14e. We use the same denominator as in part d. We can have all zinfandel, all merlot, or allcabernet, so P(all same) = P(all z) + P(all m) + P(all c) =⎛8⎞⎛10⎞⎛12⎞⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎝6⎠⎝ 6 ⎠ ⎝ 6 ⎠=⎛30⎞⎜ ⎟⎝ 6 ⎠1162593,775= .00231.a. (n 1 )(n 2 ) = (9)(27) = 243b. (n 1 )(n 2 )(n 3 ) = (9)(27)(15) = 3645, so such a policy could be carried out for 3645successive nights, or approximately 10 years, without repeating exactly the sameprogram.61