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Suppose that 30% of all Billy Lube tires are defective. If 8 tires are chosen at random, what is the probability that:

(a) Exactly 7 are defective?
(b) At least 7 are defective?

User Blahy
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1 Answer

3 votes

Final answer:

To find the probabilities, we can use the binomial probability formula: P(x) = (nCx) * p^x * (1-p)^(n-x), where n is the number of trials, x is the number of successful outcomes, p is the probability of success, and (nCx) is the number of combinations. For part (a), the probability exactly 7 tires are defective is 10.64%. For part (b), the probability of at least 7 tires being defective is 10.71%.

Step-by-step explanation:

To find the probabilities, we can use the binomial probability formula:

P(x) = (nCx) * p^x * (1-p)^(n-x)

where n is the number of trials, x is the number of successful outcomes, p is the probability of success, and (nCx) is the number of combinations.

a) To find the probability that exactly 7 tires are defective, we have n = 8, x = 7, p = 0.3:

P(7) = (8C7) * (0.3^7) * (0.7^1) = 8 * 0.3^7 * 0.7 = 0.1064 or 10.64%

b) To find the probability that at least 7 tires are defective, we need to find the probabilities of 7, 8 tires being defective and add them:

P(at least 7) = P(7) + P(8) = 0.1064 + (8C8) * (0.3^8) * (0.7^0) = 0.1064 + 0.3^8 = 0.1064 + 0.000657 = 0.107057 or 10.71%

User Gsk
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