Prob. and Stats 7th Ed. Jay L. Devore: the problem is from chapter 2, #38 (unfor

Question

Prob. and Stats 7th Ed. Jay L. Devore: the problem is from chapter 2, #38 (unfortunately, my account is messed up right now- I have to wait 24-48 hours for it to switch to "solutions," but this assignment is due tomorrow and I'm stuck on how to approach the problem; if anyone could help a bit or show the solution from that problem so I can figure it out, I'd appreciate it).

 2. A box in a certain supply room contains four 40-W light-bulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.

a. What is the probability that exactly two of the selected bulbs are rated 75 W?

b. What is the probability that all three of the selected bulbs have the same rating?

c. What is the probability that one bulb of each type is selected?

d. Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?

I think I got part b. (answer = .044), but I'm pretty lost on the rest.  Thanks for the help!

Explanation / Answer

(a) Different ways to pull two 75 watt bulbs out of the 6 that are there ---> 6! / (2! * 4!) = 15 Different ways to pull one non-75 watt bulb out of the 9 that are there ---> 9! / (1! * 8!) = 9 Multiply these two together....15 * 9 = 135. There are 135 total ways this can be done. Take 135 and divide it by 455. (135 is the total number of ways you can pull two 75-watt bulbs out of three, and 455 is the TOTAL number of ways you can choose any three bulbs.) So, this happens 135 out of 455 times. Therefore, the odds of this happening are 135/455...or 0.2967...or 29.67%.

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