In a certain shipment of 14 phones, 8 are defective. 10 of the phones are selected at random without replacement. What is the probability that 5 of the 10 phones are defective?
I'm not sure how I would go about doing this?
Prob(phone is defective) = 8/14 = 4/7
so prob(not defective) = 3/7
Prob(5 of 10 are defective) = C(10,5)(4/7)^5 (3/7)^5
= 252(1024/16807)(243/16807) = .....
This question falls in the category of
Binomial Distribution Probability
To find the probability of 5 out of 10 phones being defective, we need to use the concept of combinations.
First, let's determine the total number of possible combinations of selecting 10 phones out of the shipment of 14. This can be calculated using the formula for combinations, which is "nCr" or "n choose r," where n is the total number of items and r is the number of items chosen. The formula is:
C(n, r) = n! / (r! * (n - r)!)
For this problem, we have:
n = 14 (the total number of phones)
r = 10 (the number of phones selected)
C(14, 10) = 14! / (10! * (14-10)!)
Next, let's determine the number of combinations that have 5 defective phones out of the 8 available. We can use the same combination formula, but now:
n = 8 (the total number of defective phones)
r = 5 (the number of defective phones selected)
C(8, 5) = 8! / (5! * (8-5)!)
Finally, we can calculate the probability by dividing the number of favorable outcomes (combinations with 5 defective phones) by the total number of possible outcomes (total combinations):
Probability = C(8, 5) / C(14, 10)
Now, you can use a calculator or any software that supports factorial calculations to find the values of the combinations and then calculate the probability.