Given that, a box has 20 pens out of which 2 are defective.
Let p be the probability of a number of pens being defective out of 20 pens.
Then, q be the probability of a number of pens not being defective.
Also, p + q = 1
⇒ q = 1 – p
Let X be a random variable representing a number of defective pens out of 5 pens.
Then, the probability of getting r defective pens out of n pens is given by,
P (X = r) = nCrprqn-r
Here, n = 5
Putting all the values of n, p, and q in the above equation, we get
We need to find the probability that at most 2 pens are defective out of 5 pens.
This can be represented as,
Probability = P (X ≤ 2)
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
Put r = 0, 1, 2 in equation (i) to find P (X = 0), P (X = 1) and P (X = 2), and then substitute in the above equation. We get,
Thus, the probability that at most 2 pens are defective out of 5 pens is .
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