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What am I doing wrong here

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  • 02-10-2003 9:47am
    #1
    hussey
    Registered Users Posts: 6,240 ✭✭✭


    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


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    #2
    Ciaran
    Registered Users Posts: 915 ✭✭✭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. :)


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    #3
    hussey
    Registered Users Posts: 6,240 ✭✭✭hussey


    Now I see it


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