From a lot of 12 flares, 4 are selected at random. If the lot contains 4 defective flares *flares that will not fire), what is the probability that all 4 will not work?
What is the probability that at most 2 will not work?
To find the probability of different events, we need to understand the concept of combinations and use them to calculate the number of possible outcomes.
1. Probability that all 4 selected flares will not work:
The total number of ways to select 4 flares out of 12 is given by the combination formula: C(12, 4) = 12! / (4! * (12-4)!). This equals 12! / (4! * 8!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495.
Out of the 12 flares, 4 are defective, so the number of ways to select 4 defective flares is given by C(4, 4) = 4! / (4! * (4-4)!) = 1.
Therefore, the probability that all 4 selected flares will not work is 1/495.
2. Probability that at most 2 selected flares will not work:
First, let's find the probability that exactly 0 or 1 flare does not work:
- Probability that exactly 0 flare does not work:
The number of ways to select 4 working flares out of 8 is given by C(8, 4) = 8! / (4! * (8-4)!) = 8! / (4! * 4!) = (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) = 70.
Therefore, the probability of selecting 4 working flares is 70/495.
- Probability that exactly 1 flare does not work:
The number of ways to select 3 working flares and 1 defective flare is given by: C(8, 3) * C(4, 1) = (8! / (3! * (8-3)!)) * (4! / (1! * (4-1)!)) = (8 * 7 * 6 * 5 * 4!) / (3 * 2 * 1 * 5 * 4) = 112.
Therefore, the probability of selecting 3 working flares and 1 defective flare is 112/495.
To find the probability that at most 2 selected flares will not work, we add the probabilities obtained from the previous steps:
(70 + 112) / 495 = 182/495.
Therefore, the probability that at most 2 selected flares will not work is 182/495.