From an inventory of 10 new cars being shipped to local dealerships, corporate reports indicate that 4 have defective radios installed. If 5 cars are randomly selected for shipping, Then:
What is the probability that exactly two cars have defective radios?
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well the theoretical probability is 100% because if you divide the first group by 2 then you get 2/5,
but the real probablity is what I am thinking about
well lets see all of the possible outcomes that there will be 2 defective out of five
aaahh, its too hard I cant figure it out, all I can say is that the theoretical probability is 100%
ok go with the answerer below me, he is probably right
total number of cases = 10C5
favorable number of cases = 4C2* 6C3
Required probability = 4C2* 6C3/ 10C5
=20/21
Payel is right except the final answer is 10/21
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You can use the hypergeometric distribution to find the solution
Let X be the number of cars with a defect radio sampled. X has the hypergeometric distribution with the following parameters.
K = number of items to be drawn = 5
N = total objects = 10
M = number of objects of a given type = 4
The probability mass function for the hypergeometric distribution is defined as:
P(X = x | N, M, K) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K )
for x = {0, …, K}; M - (N - K) ≤ x ≤ K
P(X = 0 | N, M, K) = 0 otherwise
Note that the constraints on x here are very generic and it is possible to have value of K, N and M such that for x in {0, …, K} P(X = x) = 0.
If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr
the N C K is the total number of possible combinations of K objects drawn from N objects.
the M C x is the number of combinations of getting x objects of the given type
the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn.
Looking at the PMF you should be able to see that it is the ratio of the number of combination of selecting the X of the items of interest times the number of combinations of choosing K - X items from the remaining items and this is all divided by the total number of combination for choosing K items from N objects.
The expectations of the Hypergeometric distribution is KM / N = 2
The Probability Mass Function, PMF,
f(X) = P(X = x) is:
P(X = 0 ) = 0.02380952
P(X = 1 ) = 0.2380952
P(X = 2 ) = 0.4761905 ← answer
P(X = 3 ) = 0.2380952
P(X = 4 ) = 0.02380952
P(X = 5 ) = 0
The Cumulative Distribution Function, CDF,
F(X) = P(X ≤ x) is:
x
∑ P(X = t) =
t = 0
P( X ≤ 0 ) = 0.02380952380952381
P( X ≤ 1 ) = 0.2619047619047619
P( X ≤ 2 ) = 0.738095238095238
P( X ≤ 3 ) = 0.976190476190476
P( X ≤ 4 ) ≈ 1
P( X ≤ 5 ) = 1
1 - F(X) is:
K
∑ P(X = t) =
t = x
P( X ≥ 0 ) = 1
P( X ≥ 1 ) = 0.976190476190476
P( X ≥ 2 ) = 0.738095238095238
P( X ≥ 3 ) = 0.261904761904762
P( X ≥ 4 ) = 0.02380952380952395
P( X ≥ 5 ) = 1.110223024625157e-16