Please HElp Parts a to c very clear part d is confusing how

Please HElp ]]]-:

Parts \"a\" to \"c\" very clear, part \"d\" is confusing, how are terms founds, if problem had 4 (40W), 5 (60W), 7 (75W) how would part \"d\" be set up. I keep getting it wrong but don\'t understand why 1-try, 2-tries, 3-tries…??

A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected. (Round your answers to four decimal places.) 4....40-watt light bulbs 5....60-watt light bulbs 6... 75-watt light bulbs

15 total items in box Pulling three light bulbs out of 15 in the box,

there are 455 different combinations this can be done ---> 15! / (3! * 12!) = 455

(a) What is the probability that exactly 2 of the selected bulbs are rated 75 watt?

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%.

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

Different ways to pull three 40 watt bulbs out of the 4 that are there ---> 4! / (3! * 1!) = 4 Different ways to pull three 60 watt bulbs out of the 5 that are there ---> 5! / (3! * 2!) = 10 Different ways to pull three 75 watt bulbs out of the 6 that are there ---> 6! / (3! * 3!) = 20 Add these all together.... 4 + 10 + 20 = 34. There are 34 different ways this can happen. Divide 34 by 455. This equals 0.0747....or, 7.47%.

(c) What is the probability that one bulb of each type is selected. Different ways to pull one 40 watt bulb out of the 4 that are there ---> 4! / (1! * 3!) = 4 Different ways to pull one 60 watt bulb out of the 5 that are there ---> 5! / (1! * 4!) = 5 Different ways to pull one 75 watt bulb out of the 6 that are there ---> 6! / (1! * 5!) = 6 Multiply all these together.... 4 * 5 * 6 = 120. There are 120 different ways that this can happen. Divide 120 by 455. This equals 0.2637...or, 26.37%.

(d) Suppose now that the bulbs are to be selected one by one until a 75-watt bulb is found. What is the probability that it is necessary to examine at least 6 bulbs? Chances of finding it in exactly one try = 6/15 ----> 0.4 Chances of finding it in exactly two tries = 9/15 * 6/14 ----> 0.2571 Chances of finding it in exactly three tries = 9/15 * 8/14 * 6/13 ----> 0.1582 Chances of finding it in exactly four tries = 9/15 * 8/14 * 7/13 * 6/12 ----> 0.0923 Chances of finding it in exactly five tries = 9/15 * 8/14 * 7/13 * 6/12 * 6/11 ----> 0.0503

Add all these together... 0.4 + 0.2571 + 0.1582 + 0.0923 + 0.0503 = 0.9579...or, 95.79%. This is the chance that you WILL select a 75-watt bulb in the first five tries. So, the chances that it will take \"at least 6\" tries to find it is 100% MINUS 95.79%...or, 4.21%.

Solution

d)

If you need to examine at least 6 bulbs to get a 75W bulb, then that means that the first 5 bulbs are not 75W.

To get a non-75W bulb, the probability of that

For the 1st try: 9/15
For the 2nd try: 8/14
For the 3rd try: 7/13
For the 4th try: 6/12
For the 5th try: 5/11

Hence,

P(a t least 6 bulbs for 1st 75W) = (9/15)*(8/14)*(7/13)*(6/12)*(5/11)

= 0.041958042 [ANSWER]