What am I doing wrong here

hussey

Ok from the B-day thread

I tried to work out the formula on the bus this morning .. but wasn't getting anywhere

so I Tried a different technique :

3 people are asked to choose a coloured ball from a bag (5 balls in the bag) and put them back.

what is prob that at least two of them have the same colour.

Now how I used to work this out was 1- P(none having same colour)

which would be 1 - (1x 4/5x 3/5) 1- 12/25 = 13/25

now the probability for at least two is the same as the probability of them added to together
i.e. P(AB) .. probability of A and B having same colour or BC etc

P(AB) or P(AC) or P(BC) or P(ABC)
which would be 1/5 + 1/5 + 1/5 + 1/25 = 16/25

now they are different figures and I can't see where I'm going wrong!!

any one help

Ciaran

You're counting the case where they all pick the same colour four times! The cases where A and B pick the same colour include the case where they all pick the same colour. The same goes for BC and AC. As you're counting ABC seperately, you've included that possibility 4 times. 16/25-3*(1/25)=13/25.

hussey

Now I see it