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I was having a debate with a friend. The question is about probability…

If there are six people, and three will be randomly chosen, and the participants are named A, B, C, D, E, and F, what is the probability that both A and B will be chosen?

I said that since the chance for each participant to be chosen is 1/2 (6/3 = 1/2), then as laws of probability state, to find the probability of both A and B being chosen the equation can be written as 1/2 * 1/2 = x

So the probability for both being chosen should logically be 25%.

However my friend says the probability should be much lower.

I got 5 bucks betting on this anyone wanna explain this to two idiots? :D

Close. The probability is actually 20%. There are 20 possible combinations, while A and B will only be chosen together in 4 of them.

Chance of A being chosen = 3/6
Chance of B being chosen = 2/5
3/6 * 2/5 = 6/30; = 1/5 = 20%

The odds would only be 25% if the initial chosen person was replaced into the group after being chosen.

Oh, thanks a lot. This helped a lot :D

From: mixster
Chance of A being chosen = 3/6
Chance of B being chosen = 2/5
3/6 * 2/5 = 6/30; = 1/5 = 20%

The odds would only be 25% if the initial chosen person was replaced into the group after being chosen.

If the initial person would have been put back in, the chances would be 17%, because we are looking for A and B to be chosen, not twice A, or twice B.

From: mixster
Chance of A being chosen = 3/6
Chance of B being chosen = 2/5
3/6 * 2/5 = 6/30; = 1/5 = 20%

The odds would only be 25% if the initial chosen person was replaced into the group after being chosen.

confirmed. this is the correct solution.