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