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Another annoying probability question (permutations/combinations)

Sorry for the constant posts but this one is doing my head in. I can't find the answer to the 2nd part of e) a more direct way of finding the probability of being chosen for one of the parts

these are my answers so far:

a) 7P2 = 42

b) since the other position is filled there are 6 ways to fill the part for the heroine - so 6

c) 6

d) 6/42

e) let A = being chosen for the heroine

B = being chosen for the best friend

then P(A u B) = 6/42 + 6/42 = 2/7

but what is the direct route for finding this probability? I thought perhaps I could use the rule 1 minus the complement of being being chosen...but then I dont know how to calculate the probability of not being chosen!!!

Many thanks!

Re: Another annoying probability question (permutations/combinations)

Quote:

Originally Posted by

**flashylightsmeow** Sorry for the constant posts but this one is doing my head in. I can't find the answer to the 2nd part of e) a more direct way of finding the probability of being chosen for one of the parts

e) let A = being chosen for the heroine

B = being chosen for the best friend

then P(A u B) = 6/42 + 6/42 = 2/7

but what is the direct route for finding this probability? I thought perhaps I could use the rule 1 minus the complement of being being chosen...but then I dont know how to calculate the probability of not being chosen!

It really is straightforward.

If you are chosen for the heroine then you can't be the best friend and the other way round.

Thus events are disjoint and as such .

Re: Another annoying probability question (permutations/combinations)

I think I've got it!

the probability of not being chosen is found by finding the number of ways in which the 6 remaining people can be ordered (6P2 = 30) divided by the total number of ordering of the 7 people (7!)

so let A be the probability of being chosen for one of the 2 roles

then P(A) = 1-P(Ac) = 1-(30/42) = .28571

which is equal to 2/7

thats the right way of doing it....right?

Re: Another annoying probability question (permutations/combinations)

Jeez, i thought the answer was going to be more complicated than that. Thanks!