Assume any pager selected from the box is done randomly, and hence selections are independent from each other. The probability of a selected pager being defective is 0.25. Let X be the random variable for the number of defective pagers.

A) Determine the probability of selecting exactly 2 defective pagers.
B) Determine the probability of selecting at least 5 defective pagers.

I tried P(X = 2) = 0.25^2, but I'm not sure if that's the correct approach for A)

  • $\begingroup$ How many selections are made? $\endgroup$ – Graham Kemp Oct 14 '16 at 0:51
  • $\begingroup$ The question doesn't mention anything about the number of selection, so assume, that you selected 2 pagers exactly at once in A, and 5 at once in B. $\endgroup$ – Struggling Oct 14 '16 at 0:53
  • $\begingroup$ That's not really what the wording of the question suggests. ("exactly 2", "at least 5"). $\endgroup$ – Graham Kemp Oct 14 '16 at 1:03

Your approach is mkay, but really hinges on how many selections are made in total.

You are making some number, $\bbox[wheat,1pt]n$, of selections, and each with an independent rate of $0.25$ for being defection.

The count of defectives should thus have a recognisable distribution.

$$\mathsf P(X=x) = {^n\mathrm C_x}~0.25^x~0.75^{(n-x)}$$ $$\mathsf P(X\leq x) = \sum_{k=0}^x{^n\mathrm C_k}~0.25^k~0.75^{(n-k)} = 1-\mathsf P(X>x)$$

  • $\begingroup$ So I plugged n = 6, because that's the number of samples, and x = 2. I get 35% as the chance of exactly two items selected being defective. Does this make sense? Or am I doing it wrong? $\endgroup$ – Struggling Oct 14 '16 at 1:11
  • $\begingroup$ @Struggling Not sure how you are getting that. $$\dfrac{6!}{2!~4!}\dfrac{1^2\,3^4}{4^6} = \dfrac{1215}{4096} = 0.296630859375$$ $\endgroup$ – Graham Kemp Oct 14 '16 at 1:53

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