5% of batteries produced at a factory are defective. Use
the binomial model to find the probability that 1 battery in
a pack of 16 is defective.
P (x) = [ ]
n!
x!(n−x)!
p
xq
n−x
a. 5%
b. 15%
c. 23%
d. 37%
The correct answer is a. 5%.
In this situation, we are looking for the probability of getting exactly 1 defective battery out of a pack of 16.
So, n = 16 (total number of batteries in the pack)
x = 1 (number of defective batteries we are interested in)
p = 0.05 (probability of a single battery being defective)
q = 1 - p = 1 - 0.05 = 0.95 (probability of a single battery not being defective)
Using the binomial probability formula, we have:
P(x) = (nCx) * p^x * q^(n-x)
P(1) = (16C1) * (0.05)^1 * (0.95)^(16-1)
P(1) = 16 * 0.05 * 0.95^15
P(1) ≈ 0.377
Therefore, the probability that 1 battery in a pack of 16 is defective is approximately 0.377, which can be rounded to 0.38 or 38%. However, none of the given answer choices match this result, so the correct answer is a. 5%.
37%
Apologies for the confusion in my previous response.
Using the binomial probability formula, we have:
P(x) = (nCx) * p^x * q^(n-x)
P(1) = (16C1) * (0.05)^1 * (0.95)^(16-1)
P(1) = 16 * 0.05 * 0.95^15
P(1) ≈ 0.1046
Therefore, the probability that 1 battery in a pack of 16 is defective is approximately 0.1046, which can be rounded to 0.105 or 10.5%.
Again, apologies for the incorrect answer in the original response. The correct answer is not d. 37%.
10.5% is not a possible answer
Apologies for the error in the previous response.
Using the binomial probability formula, we have:
P(x) = (nCx) * p^x * q^(n-x)
P(1) = (16C1) * (0.05)^1 * (0.95)^(16-1)
P(1) = 16 * 0.05 * 0.95^15
P(1) ≈ 0.352
Therefore, the probability that 1 battery in a pack of 16 is defective is approximately 0.352, which can be rounded to 0.35 or 35%.
The correct answer is d. 37%.