# Thread: Probability problem

1. ## Probability problem

I was hoping someone could check my work as something seems off to me. Any help/suggestions would be greatly appreciated.

(a)From a lot of 12 flares, 4 are selected at random. If the lot contains 4 defective flares, what is the probability that all 4 will not work?

P(4 defective)= C(4,4) / C(12,4)
= 1/149
= 0.002 OR 0.2%

(b)What is the probability that at most 2 will not work?
P(at most 2 will not work) = P(0, 1 and 2 will not work)

P (0 will not work) = (C(4,0) * C(8,4)) / C(12,4) = 0.1414...

P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131

P (2 will not work) = (C(4,2) * C(8,2)) / C(12,4) = 0.3394

Total = 0.5924 or 59.24%

Thank you for viewing/responding

2. Originally Posted by jake.davis
I was hoping someone could check my work as something seems off to me. Any help/suggestions would be greatly appreciated.

(a)From a lot of 12 flares, 4 are selected at random. If the lot contains 4 defective flares, what is the probability that all 4 will not work?

P(4 defective)= C(4,4) / C(12,4)
= 1/149
= 0.002 OR 0.2%

(b)What is the probability that at most 2 will not work?
P(at most 2 will not work) = P(0, 1 and 2 will not work)

P (0 will not work) = (C(4,0) * C(8,4)) / C(12,4) = 0.1414...

P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131

P (2 will not work) = (C(4,2) * C(8,2)) / C(12,4) = 0.3394

Total = 0.5924 or 59.24%
Thank you for viewing/responding
I have not done the calculations.
But the methods are correct.

3. I took a look at some of your calculations and i saw that for part b.
P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131
when i calculated it i came up with this, i wrote it out so you can see...
P = 4*56/ 495
= 224 / 495
= 0.4525....
other than that one part the rest of your calculations are correct as far as i've calculated!