probability exercise with dependent events

I have some difficulties in solving the following problem:

A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events E1, E2, E3, E4 as follows:

E1 = {the first pile has exactly 1 ace},

E2 = {the second pile has exactly 1 ace},

E3 = {the third pile has exactly 1 ace},

E4 = {the fourth pile has exactly 1 ace},

Use the formula

$\displaystyle P(E_{1}\cap E_{2}...E_{n})=P(E_{1})P(E_{2}\mid E_{1})P(E_{3}\mid E_{1}\cap E_{2})...P(E_{n}\mid E_{1}...E{n-1}) $

to find

$\displaystyle P(E_{1}\cap E_{2}\cap E_{3}\cap E_{4})$

the probability that each pile has an ace.

Re: probability exercise with dependent events

Looks like the product of four hypergeometrics

Quote:

Originally Posted by

**achacy** I have some difficulties in solving the following problem:

A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events E1, E2, E3, E4 as follows:

E1 = {the first pile has exactly 1 ace},

E2 = {the second pile has exactly 1 ace},

E3 = {the third pile has exactly 1 ace},

E4 = {the fourth pile has exactly 1 ace},

Use the formula

$\displaystyle P(E_{1}\cap E_{2}...E_{4})=P(E_{1})P(E_{2}\mid E_{1})P(E_{3}\mid E_{1}E_{2})P(E_{4}\mid E_{1}E_2E_3) $

to find

$\displaystyle P(E_{1}\cap E_{2}\cap E_{3}\cap E_{4})$

the probability that each pile has an ace.

$\displaystyle P(E_{1})= {{4\choose 1}{48\choose 12}\over {52\choose 13}} $

That's how the first pile has exactly one ace and the rest has three.

So, now we have 3 aces and 39 total cards, so

$\displaystyle P(E_{2}\mid E_{1})= {{3\choose 1}{36\choose 12}\over {39\choose 13}}$

That leaves us with 2 aces and 26 cards...

$\displaystyle P(E_{3}\mid E_{1}E_2)= {{2\choose 1}{24\choose 12}\over {26\choose 13}}$

Finally we are stuck with a fourth pile that MUST have one ace...

$\displaystyle P(E_{4}\mid E_{1}E_2E_3)= {{1\choose 1}{12\choose 12}\over {13\choose 13}}$

The product of these probabilities should be the answer.

Re: probability exercise with dependent events

Thanks a lot for this help. It seems that I have to refresh my knowledge of combinatorics...

Re: probability exercise with dependent events

Thats just the basic hypergeometric, which is my next lecture in Probability on Tuesday.