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Same as other question but add together exactly 0, exactly 1 and exactly 2

0 = 6/252
1 = 60/252
2 = 120/252

no more than 2 = 186/252 = 31/42

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2

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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
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 ← answer
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