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**iamstuck****Member**- Registered: 2005-07-06
- Posts: 1

please can somebody help me understanding the maths to work out the probability of :-

if i had eight identical pieces of paper, 2 of which had the letter A written on them, 2 with B, 2 with C

and 2 with D written on them , jumbled up upside down on the table . if i was to randomly arrange them into pairs what would the probability be of ending up with AA BB CC DD ?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Randomly pick one. The remaining papers will have one matching letter and three pairs of different letters. This means that for the first pair there is a 1 in 7 chance of them matching.

For the second pair, the same reasoning applies except that there are now only two pairs of different letters so there is a 1 in 5 probability of them matching.

The third pair has a 1 in 3 chance and if all 3 of these pairs match then the last pair must also match.

The total probability is worked out by multiplying all of these together, so it is 1 in (7x5x3x1) or if you want to be really technical, 1 in 7!!, which works out to be 1 in 105.

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You did your 5th post.

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You've done at least 30 posts.

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I'm not sure, but looking at the user list tells you that it's somewhere between 196 and 286.

I'm guessing 200.

Why did the vector cross the road?

It wanted to be normal.

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