In a book called Applied Statistics and Probability for Engineers (Fourth edition) by Douglas C. Montgomery and George C. Runger there is a problem that reads: "A lot of 100 semiconductor chips contains 20 that are defective. Two are selected randomly, without replacement, from the lot.
(a) What is the probability that the first one selected is defective ?
(b) What is the probability that the second one is defective given that the first one was defective ?
(c) What is the probability that both are defective ?
My solution
Let the set $ A $ be the one that contains all the 100 semiconductor chips and $ B $ the set that contains all the chips except the one we selected firstly. The sample space of our experiment is:
$$ S = A \ \times B = \{(x,y):x \in A \ \land y \in B\}$$ The number of possible outcomes is $$ |S|=|A|\cdot|B|=100 \cdot 99=9900 $$
For question (a): Let $ C $ be the set of the defective semiconductor chips that are defective. The event of interest is:
$$ K = \{(x,y)\in S:x\in C \ \land y\in B \}=C \times B $$ The probability of $ K $ is (assuming uniform probability measure): $$ \Bbb P(K)=\frac{|K|}{|S|}=\frac{|C| \cdot|B|}{|A| \cdot |B|}=\frac {|C|}{|A|}=0.2 $$ For question (b): Let $ U $ be the set that contains all the elements of S where the second selected item is defective. The probability of interest is: $$ \Bbb P(U|K)= \frac{\Bbb P(U \cap K)}{\Bbb P(K)} $$ We know $ \Bbb P(K) $ but we don't know the value of the numerator which asks us to find the the probability that both choices are defective. That probability is just: $$ \Bbb P(U\cap K)=\frac{|20| \cdot |19|}{9900} $$ Finally, we get: $$ \Bbb P(U|K) = 0.19 $$ For part (c): We found in part (b) that $ \Bbb P(U \cap K)= 0.038 $
Questions
First of all, the problem asks us to find the joint probability of $ U $ and $ K $ after question (b) which asks for the conditional probability of $ U $ given $K $. The thing is that, in order to calculate that probability we need to find the probability of the $ \Bbb P (U \cap K) $ as stated in the solution. I feel that there is a reason the authors asked the questions in this specific order. So, Is there a way of calculating $ \Bbb P (U|K) $ without knowing $ \Bbb P(U \cap K)$ ?
Secondly, I always approach probability problems by defining sets and working with their cardinalities. This method is safe but slow (not good for exams). Are there any faster methods ?
Thank you in advance for comments / answers.
Here is how I would do it.
a) Any chip is equally likely to be selected, so the probability of choosing a defective one is $${20\over100}$$ the number of defective chips divided by the total number of chips.
b) Imagine that a defective chip has been chosen. Then there are $19$ defective chips and $99$ total chips remaining, so the conditional probability is $${19\over99}$$ for the same reason as in part a).
c) The probability that both are defective is the product of the probabilities in parts a) and b) or $${20\cdot19\over100\cdot99}$$ As a a check this is the same answer we would get if we divided the ${20\choose2}$ ways to choose two defective chips by the ${100\choose2}$ way to choose two chips, which is the way I would do part c) if parts a) and b) hadn't been asked.