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There are 10 units, 4 of which are defective. 5 of the 10 are randomly selected, and I'm looking for the probability that at least 3 of the defective units are included in that selection.

To solve this I found the probability of getting 3 defective units, then the probability of getting 4 defective units (since those are the only options), and added them together. Thus, the total probability is: [(4c3)(6c2)+(4c4)(6c1)]/(10c5) Where 4c3 means "4 choose 3"

Is this correct? I'm not sure how to determine whether or not adding the probabilities is allowed in this case. Also, is there another method to solving this problem?

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  • $\begingroup$ I will never understand why it is not simply $0.6^2 0.4^3$ $\endgroup$
    – N74
    Aug 31, 2016 at 20:47
  • $\begingroup$ @N74 Probably because I'm not comfortable enough with the material to process it outside of counting rules yet. $\endgroup$
    – IgnorantCuriosity
    Aug 31, 2016 at 20:53
  • $\begingroup$ @N74 $0.6^20.4^3$ is the probability of extracting, with replacement, two okay units and then three defective units in that order. $\endgroup$
    – Graham Kemp
    Aug 31, 2016 at 20:55
  • $\begingroup$ @grahamkemp... I can't multply by $5!$ :) $\endgroup$
    – N74
    Aug 31, 2016 at 21:01

1 Answer 1

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Yes; you have it.   To confirm:

You want the probability that at least three units are defective in a sample of five units from a population of four defective and six okay.   (A sample is drawn without replacement.)

In this case, "at least 3" is either "exactly 3" or "exactly 4".   That is why you added.   (It is the additive law for the probability of a union of disjoint events.)

Let $X$ be the count of defective in the sample.   This is a hypergeometricaly distributed random variable.

$$\begin{align}\mathsf P(X\geq 3) ~=~& \mathsf P(X=3)+\mathsf P(X=4)\\[1ex] =~& \dfrac{{^4{\rm C}_3}{^6{\rm C}_2}+{^4{\rm C}_4}{^6{\rm C}_1}}{{^{10}{\rm C}_5}}\end{align}$$

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