that is probabilty question ,a challenge?

Question

suppose i had to find two light bulbs from collection of 12 but all of three which are working.if i test each one in turn what is probability that i would find two working bulbs.
a:among the first three bulbs tested
b:when three bulbs have been tested,but not beore

suppose i had to find two light bulbs from a collection of 12,all but 3 of which are working.if i test each one in turn,what is probability that i would find two working bulbs
a:among the first three bulbs tested
b:when three bulbs have been tested,but not before?

Answer 1

a) probability of finding exacttly 2 working bulbs in first three = 3*( 9/12*8/12*3/10) = 27/55.


b)
probability of not finding even one = 3/12 * 2/11 * 1/10 = 1/220
probability of finding exactly 1 = 3 * ( 9/12 * 3/11*2/10) = 9/220

thus required probability = 10/220 = 1/22.

Answer 2

There are 9 bulbs that work "O" and 3 that don't "X".

a) When testing the 3 bulbs there are 3 ways of choosing exactly 2 working bulbs out of the 3: OOX, OXO, XOO

The probability for each of these 3 cases is:

(9/12)*(8/11)*(3/10) = 9/55 = 0.163636....

The total probability of exactly 2 out of first 3 working is:

3 x 9/55 = 27/55 or about 49.1%

b) This will happen if you have OXO or XOO. So the probabilty of this is just 2 x 9/55 = 18/55 or about 32.7%

Note: you have to be careful how you word these problems, since the probability that the first 2 bulbs will work (OO) = 9/12 x 8/11 = 6/11 or about 54.5%. If you quit testing because you found two working bulbs then it is a different situation. You would be calculating the odds of getting exactly 2 working bulbs in two or three tests, and the probability of this is greater and equals:

6/11 + 18/55 = 48/55 = about 87.3 %

And you could also have all three bulbs working OOO and the probability of this is 6/11 x 7/10 = 21/55 about 38.2%

Answer 3

A. The easiest procedure is to compute the number of ways you could fail to get two or more good bulbs, then subtract from 1. If you draw 3 bulbs randomly, there is only 1 combination that will result in zero working bulbs (because there are only 3 bad bulbs to begin with). There are 27 combinations of only one working bulb: bad bulbs A and B with any of 9 good bulbs; bad bulbs B and C with any of 9 good bulbs; and bad bulbs A and C with any of the 9 good bulbs. So there are only 27 + 1 = 28 ways to choose less than two working bulbs.

The total number of combinations of 3 drawn from a group of 12 is: 12! / 6 = 79,833,600. So the test fails 28 times out of almost 80 million, or 1 time in 2,851,200 to be precise. So you will find two working bulbs 2,851,199 times out of 2,851,200 tries.

B. To succeed at this test, exactly one of the first two draws must be a bad bulb, and the third draw must be a good bulb.

B1. Probability that the first draw will be bad: 3 / 12 or 1/4. IFF bad bulb is drawn, probability that the next bulb will be good: 9 / 11. Probability of BOTH events occurring: 3/12 x 9/11 = 9 / 44.
B2. Probability that the first draw will be bad: 9/12 or 3/4. IFF good bulb is drawn, probability that next draw will be bad: 3/11. Probability of BOTH events occurring: 3/4 x 3/11 = 9/44.
B3. Probability that EITHER B1 OR B2 will occur: 9/44 + 9/44 = 18/44.
B4. Probability of drawing a bad bulb given that B3 has occurred: 2/10 or 1/5. Probability of BOTH B3 and B4 occurring is 18/44 x 1/5 or 18/220.

Answer 4

This sounds like a beautiful problem in probability but I don't quite understand the conditions. Could you rephrase it, please? For example, are you assuming you know beforehand that there are exactly three light bulbs in wording conditions in the group of 12?
Also, if by chance the first two bulbs tested work properly, you wouldn't have to test a third, would you?

Answer 5

Can you check the filament to see if it is broken? This will increase the odds in your favor if filament is visible.

Answer 6

a) - - - - - - -6/132