Probability confusion

[kimd6746]

I'm really getting stuck on an OG question so I have rephrased to the following:

Team A has 500 members. Team B has 200 members. Among these members there are 20 siblings pairs, each consisting of one member from each team. If one member is to be selected at random from each team, what is the probability that the 2 members selected will be a sibling pair?

I was thinking that since there are 60 siblings "pairs," there are 60 from team A and 60 from team B. Therefore, I concluded that the probability of picking a sibling pair from team A is 60/500.

Then I thought the probability of picking the other sibling pair from team B would be 60/200.

Therefore my answer should be (60/500) x (60/200) = 9/250

But this is incorrect! Why? The correct answer is (60/500) x (1/200) = 3/5000.

Why must the 2nd event be 1/200 when the "pair" suggests that there are 60 siblings in team A and 60 in team B?

[Guest]

I'm really getting stuck on an OG question so I have rephrased to the following:

Team A has 500 members. Team B has 200 members. Among these members there are 20 siblings pairs, each consisting of one member from each team. If one member is to be selected at random from each team, what is the probability that the 2 members selected will be a sibling pair?

I was thinking that since there are 60 siblings "pairs," there are 60 from team A and 60 from team B. Therefore, I concluded that the probability of picking a sibling pair from team A is 60/500.

Then I thought the probability of picking the other sibling pair from team B would be 60/200.

Therefore my answer should be (60/500) x (60/200) = 9/250

But this is incorrect! Why? The correct answer is (60/500) x (1/200) = 3/5000.

Why must the 2nd event be 1/200 when the "pair" suggests that there are 60 siblings in team A and 60 in team B?

I'm confused. How did you get 60 siblings from a question you wrote that said "20 pairs" ?



If there were 60 sibling pairs (120 people) then Team A would have 60 out of 500 (60/500) . Your chance of getting one of these guys is 12%.

The next team has 200 members on it and we're trying to find only 1 member because 199 out of the 200 aren't related to our guy. (1/200)

This is an "AND" probability case so you multiply the probabilities. 1/200 x 60/500

[kimd6746]

Oops, sorry! You're right, I mean to to say 60 pairs from the get go!

Ok, your explanation makes total sense! I think I outsmarted myself. If there is a 12% chance of picking a sibling pair, then obviously there can be only one person available to choose from Team B because who else can it be other than their sibling half! D'oh!!!!

[rfernandez]

Always good to see folks helping each other out. Nice work.

[Guest]

Shouldn't the ans in this case be 3/2500 since we have two cases here.

1) 60/200*1/500 = 3/5000

2) 60/500*1/200 = 3/5000 since I can pick a member from the 200 group first and then 500 or first from 500 and then 200.

[rfernandez]

Shouldn't the ans in this case be 3/2500 since we have two cases here.

1) 60/200*1/500 = 3/5000

2) 60/500*1/200 = 3/5000 since I can pick a member from the 200 group first and then 500 or first from 500 and then 200.

But these two calculations refer to the same case: 1 member from each team and both are related to each other.

Instead, you found two different ways to calculate the same probability.

[cramya]

Not sure if I am re-explaning a solved problem. But here's my 2 cents worth.

The probability is 60/500 * 1/200 or 60/200*1/500. The reason its 1 and not 60 on the other side of the multiplication here is because once one of the sibling pair is picked from one team there is only one other member in the other team that would be the corresponding sibling

[RonPurewal]

Shouldn't the ans in this case be 3/2500 since we have two cases here.

1) 60/200*1/500 = 3/5000

2) 60/500*1/200 = 3/5000 since I can pick a member from the 200 group first and then 500 or first from 500 and then 200.

nope.

you SHOULD take into account OUTCOMES THAT ARE ACTUALLY DIFFERENT.
you should NOT take into account DIFFERENT METHODS FOR FINDING THE SAME OUTCOMES; you should just choose one method and stick with it.

here's an analogy:
let's say that you have 3 shirts and 4 pairs of pants. hopefully, you know that this makes 12 outfits (at least if you aren't conscious of style, anyway...)
INVALID ARGUMENT:
"well, i could pick a shirt first and then a pair of pants, so that would be 3 x 4 = 12 outfits, but i could also pick a pair of pants first and then a shirt, so that would be 4 x 3 = 12 more outfits. therefore, there are 24 outfits total."
you can see what's wrong with this argument, right? you aren't actually counting 12 more outcomes; you're just counting the same 12 outcomes that you already found, only using a different criterion. that's bad: now you're double-counting each of the outfits.

the same is happening in your derivation here.
hth

[san]

That is the diffrence between permutation and combination.

[JonathanSchneider]

Well, not really, San. What Ron is driving at is that there are multiple ways for solving for a particular number. You don't want to go ahead and then add together all of these values, just because you found new ways to solve the problem. This is the same for permutations as for combinations or any other type of problem.