I would make a tree diagram for this one, have you tried that?
All the probability i know is from intuition which is why I'm stuck on a question as simple as this:
In a sample of 100 people 20% are tall (over 6 feet in height) and 10% are wearing a blue shirt, what is the probability that a person chosen at random is tall and wearing a blue shirt?
At first i thought there was no solution because it could be that none of the tall people are wearing a blue shirt but then I thought what if you had to calculate the probability of choosing a tall person with a blue shirt in every possible arrangement and then take the average. Of course the 2nd option is too time consuming for a problem like this and unlikely to be the actual answer, but I thought maybe there is a quick calculation that yields the same results as option 2.
BTW I made this problem, but I saw one almost exactly like it online (not from a very credible source) so it's very possible that there is no solution (from my perspective).
So that is 10% of 20% of the population are both tall and wearing blue shirts, so on average from your (random) sample of 100 people 20 are tall and 2 of those are wearing blue shirts (2 is 10% of the 20 talls) .
In practice you will not always find exactly two tall blue shirt wearers but may find 0, 1, 2, 3, 4, ...
If you wish to pursue this further the number "n" of tall blue shirt wearers in a random sample of 100 has a Binomial distribution b(n,100,0.02) (use Google/Wikipedia for more information).
You have 4 writing utensils, 3 of them are pens (75%) and 2 of them work (50%). Now using your method there are 1.5 working pens, but if you write out all the possible combinations, take all the chances of getting a working pen and average them together you get 45% or 1.8 pens.
EDIT: it seems that assuming that a property that pertains to a group also pertain in to the same extend to a subgroup of that group leads to false results, is it method actually used in higher level probability?
Let A,B be two events, given that they are independent then P(AnB)=P(A)*P(B)
here in your first question,
A ----> no of persons who are tall
B ----> no of persons who wear blue shirt
You clearly know that if a person is tall it has nothing to do with whether he wears a blue shirt or not,so A & B are independent events,
so,we can apply the above formula,
P(AnB) = (2/10)(1/10) = 0.02
it implies out out of 1 there is a chance of 0.02,
so for 100 people 2 May be tall with blue shirts
Also your problem is now worded in deterministic terms, It for the argument I presented earlier to work it should be worded: "We choose a random sample of 4 writing utensils from a population where 75% of the utensils are pens and independently 50% of the utensils work. What is the expected number of working pens in our sample?"
However you will get the same numerical answer if you have exactly 3 of the four implements are pens and exactly 2 of the implements work and now enumerate the cases.