chapter 1

Chapter 2: Probabilityd. To examine exactly one, a 75 watt bulb must be chosen first. (6 ways to accomplish this).To examine exactly two, we must choose another wattage first, then a 75 watt. ( 9 × 6ways). Following the pattern, for exactly three, 9 × 8 × 6 ways; for four, 9 × 8 × 7 × 6;for five, 9 × 8 × 7 × 6 × 6.P(examine at least 6 bulbs) = 1 – P(examine 5 or less)= 1 – P( examine exactly 1 or 2 or 3 or 4 or 5)= 1 – [P(one) + P(two) + … + P(five)]⎡ 6 9 × 6 9 × 8 × 6 9 × 8 × 7 × 6 9 × 8 × 7 × 6 × 6 ⎤= 1−⎢ + + ++⎥⎣1515 × 14 15 × 14 × 13 15 × 14 × 13×12 15 × 14 × 13 × 12 × 11⎦= 1 – [.4 + .2571 + .1582 + .0923 + .0503]= 1 - .9579 = .042139.a. We want to choose all of the 5 cordless, and 5 of the 10 others, to be among the first 10⎛5⎞⎛10⎞⎜ ⎟⎜⎟⎝5⎠⎝5 ⎠⎛15⎞⎜ ⎟⎝10⎠2523003serviced, so the desired probability is = = . 0839b. Isolating one group, say the cordless phones, we want the other two groups represented inthe last 5 serviced. So we choose 5 of the 10 others, except that we don’t want to includethe outcomes where the last five are all the same.⎛10⎜⎝ 5So we have⎛⎜⎝⎡⎛10⎞⎤3⋅⎢⎜⎟ − 2⎥⎣⎝5 ⎠ ⎦=⎛15⎞⎜5⎟⎝ ⎠⎞⎟ −⎠15⎞⎟5 ⎠2. But we have three groups of phones, so the desired probability is3(250)3003= .2498.c. We want to choose 2 of the 5 cordless, 2 of the 5 cellular, and 2 of the corded phones:⎛5⎞⎛5⎞⎛5⎞⎜ ⎟⎜⎟⎜⎟⎝2⎠⎝2⎠⎝2⎠=⎛15⎞⎜ ⎟⎝ 6 ⎠10005005= .199865