For this question: A fair 6-sided die is rolled five times and a fair coin is flipped three times. What is the probability of rolling an even number in all five rolls of the die and flipping heads in all three coin flips?
Why was combinations not used to arrange these events?
Can someone explain me this?
Because if I told you (x,y) = (1,2) then you don’t consider its permutations, do you? (E_1, E_2) = (P_1, P_2) where E_i is the corresponding event and P_i is the corresponding probability.
I didnt understand can you explain with some examples
What did you not get? Sadly, there’s not much examples to this besides what I already gave you.
The probability of the first even is A and the probability of the second even is B, why would you start arranging them? For example, you wouldn’t permute (x,y) = (1,2) and consider something like (y,x) = (2,1) because they’re practically the same thing, no?
Btw i wrote a more comprehensive answer there for the same question, but it’s a bit more math-intensive. You can read it if you like
Combinations because either heads can occur first or die can. And internally they can arrange in their respective possible ways. I observed your explanation still I am unable to understand.
Well if you toss two coins then the sample space of (C_1, C_2) is as follows:
\{ (H,H), (T,H), (H,T), (T,T) \}
Did it matter if coin 1 was flipped first or second? We just know that for each tuple in the sample space, the first element corresponds to C_1 and the second element corresponds to C_2. There’s no permutation/combination going on here because the order in which coin is tossed first doesn’t have anything to do with something like computing the probability of flipping two heads. If coin 1 was tossed first then you get 1/4 this is invariant if coin 2 was tossed first.
In general, we are distinguishing coins with each coin flip having its own random variable. Owing to that, we’re looking for a tuple which gives us exactly the values of the random variable we’re concerned with. In this case, we’d be looking for C_1 = H and C_2 = H and there’s no permutations to consider in that process.