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.
Probability and Statistics - Reiny, Sunday, September 29, 2013 at 8:14am
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?
I used [4!/(2!2!)](.9^3)(.1^2) for a)
Do I have the right formula for part b)? a probability of .99117 does not seem to make sense.
Katie, you were right, there should have been
3 successes and 2 failures, which would be
6( .9^3)(.1^2)
Don't know why I miscounted.
in my "prob( a specific 2F,2S, with S at end) = (.9^2)(.1^2)(.1) = .00081 "
I had the first part correct, with 2F's and 3S's, but somehow put in .1 instead of .9
b) your formula seems right
could be 3 S's OR 3 S's 1 F OR 3 S's 2 F , in each case S at the end
= .9^3 + 3 (.9^2) (.1) + 6(.9^3)(.1^2)
= .729 + .192 + .04374
= .96474
you probably just made a calculation error
Well, let's take a closer look at part b.
To find the probability that the third nondefective engine will be found on or before the fifth trial, we need to consider all the possible outcomes.
The possible outcomes for the first four trials could be either: SFFFS or SFFSF or SFFF
Let's calculate the probabilities for each of these outcomes:
For SFFFS: (0.9)(0.1)(0.1)(0.1)(0.9) = 0.000081
For SFFSF: (0.9)(0.1)(0.1)(0.9)(0.9) = 0.000729
For SFFF: (0.9)(0.1)(0.1)(0.9)(0.9) = 0.000729
Now, let's add up these probabilities:
0.000081 + 0.000729 + 0.000729 = 0.001539
So, the probability that the third nondefective engine will be found on or before the fifth trial is approximately 0.001539.
I hope this clarifies things for you!
Yes, you used the right formula for part a). The probability that the third nondefective engine will be found on the fifth trial is calculated using the negative binomial distribution.
For part b), you need to find the probability that the third nondefective engine will be found on or before the fifth trial. This means that it could be found on the first, second, third, fourth, or fifth trial.
To calculate this, you can add up the probabilities of finding the third nondefective engine on each of these trials:
P(X <= 5) = P(X = 3) + P(X = 4) + P(X = 5)
Using the negative binomial distribution formula, you can calculate each of these probabilities and sum them up.
To solve part b) of the problem, we need to find the probability of finding the third nondefective engine on or before the fifth trial.
We can approach this problem by considering the complementary event, which is finding the third nondefective engine after the fifth trial.
The probability of finding the third nondefective engine after the fifth trial can be calculated using the negative binomial distribution. The formula for the probability mass function of the negative binomial distribution is:
P(X = k) = (k-1)C(r-1) * p^r * (1-p)^(k-r)
Where:
- X is the random variable representing the number of trials until the rth nondefective engine is found,
- k is the number of trials,
- r is the number of successes (nondefective engines),
- p is the probability of success (finding a nondefective engine), and
- C(n, r) is the number of combinations of n items taken r at a time.
In this case, we want to find the probability of finding the third nondefective engine on or before the fifth trial, which is the complement of finding it after the fifth trial.
Using the formula for the negative binomial distribution and summing up the probabilities for k = 3, 4, and 5, we can calculate the probability of finding the third nondefective engine on or before the fifth trial.
P(X ≤ 5) = P(X = 3) + P(X = 4) + P(X = 5)
= (2C2 * 0.9^3 * 0.1^2) + (3C2 * 0.9^4 * 0.1^1) + (4C2 * 0.9^5 * 0.1^0)
= (1 * 0.9^3 * 0.1^2) + (3 * 0.9^4 * 0.1) + (6 * 0.9^5)
Calculating this expression will give us the correct probability for part b) of the problem.