A batch of 50 different automatic typewriters contains exactly 10 defective machines. What is the probability of finding:

(a) At least one defective machine in a random group of 5 machines?

(b) At least two defective machines in a random group of 10 machines?

(c) The first defective machine to be the kth machine taken apart for inspection in a random sequence of machines?

(d) The last defective machine to be the kth machine taken apart?

I believe that finding at least two defective machines would be :

P(x≥2) = 1 − P(x<2) = 1 − P(x=0) − P(x=1)

and similarly for at least one defective machine (as in part (a)). I'm not sure whether this would be correct or not and I have no idea how to do parts (c) or (d) at all.

  • $\begingroup$ Here's a hint: There is a 1 in 5 chance of getting a defective machine. Think of a machine selection as a roll of a 5 sided die where a 1 is a defect. Translate your questions above into the language of rolling 5 sided dice and looking for 1s and the problem should get easier. $\endgroup$ – John Douma Sep 14 '15 at 16:00
  • $\begingroup$ Here, the probabilities do not remain constant from trial to trial. $\endgroup$ – true blue anil Sep 14 '15 at 17:31

Note, this is selection without replacement so the hypergeometric distribution should be used.

Your approach for parts (a) and (b) are ok

For (c), you need $(k-1)$ ok machines followed by a defective one, so you could compute

$$\left(\frac{40}{50}\right)\left(\frac{39}{49}\right)\left(\frac{38}{48}\right). . .\left(\frac{40 - k+2}{50-k+2}\right)\left(\frac{10}{50-k+1}\right)$$

For (d), note that you must have 9 defectives in the first (k-1) followed by the last defective, whereas randomly they could be anywhere in the 50, so

$$\frac{{k-1\choose 9}{1\choose1}}{50\choose10} = \frac{k-1\choose 9}{50\choose10}$$


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