Assume 10% of the engines manufactured on an assembly line are defective. If engines are randomly selected one at a time and tested. What is the probability that the third nondefective engine will be found
a)on the fifth trial?
b) on or before the fifth trial?
MY answers:
a) P(X=5)=.0437 (Used the negative bin. dist)
b)P(X<=5)=P(X=3)+P(X=4)+P(X=5)=.99117
I think b) is definitely wrong. I used the negative binomial distribution to get all of the probabilities, I don't know what else to do for b.
If I understand correctly, there are 5 trials and you are expecting 3 S's and 2 F's, (S = success, F = failure)
defective - F -----> prob(F) = .1
non-defective - S --> prob(S) = .9
a) you want an S in the 5th spot
e.g. SFFSS is one of these
number of ways for your specified event (having an S at the end) = C(4,2) x C(2,2) = 6
or 4!/(2!2!) x 1 = 6
prob( a specific 2F,2S, with S at end) = (.9^2)(.1^2)(.1) = .00081
prob of your event = 6(.00081) = .0486
mmmh, close to your answer.
Did you interpret this differently?
To calculate the probabilities, we need to understand the concept of the negative binomial distribution. The negative binomial distribution represents the number of independent and identically distributed Bernoulli trials required to obtain a fixed number, "r," of successes.
Let's solve both parts of the problem step by step:
a) Probability of finding the third non-defective engine on the fifth trial:
To find the probability that the third non-defective engine is found on the fifth trial, we can calculate the probability that the first four engines are not defective, and the fifth one is not defective.
The probability of finding a non-defective engine on any single trial is 1 - 0.10 (10% defective rate) = 0.90.
The probability of finding a non-defective engine on the fifth trial can be calculated as follows:
P(X=5) = P(4 non-defective engines before the fifth trial) * P(non-defective engine on the fifth trial)
= (0.90)^4 * 0.90
= 0.90^5
≈ 0.59049
Therefore, the probability that the third non-defective engine will be found on the fifth trial is approximately 0.59049.
b) Probability of finding the third non-defective engine on or before the fifth trial:
To find the probability that the third non-defective engine is found on or before the fifth trial, we need to calculate the cumulative probability up to the fifth trial.
P(X≤5) = P(X=3) + P(X=4) + P(X=5)
To find these probabilities individually, we can use the negative binomial distribution formula:
P(X=k) = (r-1)C(k-1) * p^(r) * (1-p)^(k-r), where C denotes combination.
P(X=3) = (2)C(2) * 0.90^(3) * (0.10)^(2) = 1 * 0.90^3 * 0.10^2 = 0.0729
P(X=4) = (3)C(3) * 0.90^(4) * (0.10)^(3) = 1 * 0.90^4 * 0.10^3 = 0.06561
P(X=5) = (4)C(4) * 0.90^(5) * (0.10)^(4) = 1 * 0.90^5 * 0.10^4 = 0.059049
P(X≤5) = 0.0729 + 0.06561 + 0.059049 ≈ 0.197559
Therefore, the probability that the third non-defective engine will be found on or before the fifth trial is approximately 0.197559.
Your answer for part a) is correct (P(X=5) ≈ 0.59049), and for part b), the correct cumulative probability is approximately 0.197559.