1. ## Probability question

a deck of cards is shuffled and then divided into two halves of 26 card each. A card is drawn from one of the halves, it turns out to be an ace. the ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.
Hint: condition on whether or not the interchanged card is selected.

the answer is 129/(51*27) and I don't understand why.

2. ## Re: Probability question

$P[\text{pick ace from 2nd deck}] = P[\text{pick the ace from the first deck}] + P[\text{pick an ace originally in second deck}]$

The left hand term is simple. There are 27 cards in the second deck after inserting the ace from the first deck. The probability you pick that ace from the first deck is just

$P[\text{pick the ace from the first deck}] = \dfrac{1}{27}$

The right hand term is trickier and depends upon the number of aces that were originally in the 2nd deck to start with. It can be 0-3, as we know that 1 was in the first deck.

$P[\text{pick an ace originally in second deck}] = \displaystyle \sum_{k=0}^3~P[\text{pick an ace originally in second deck | k aces in second deck}]P[\text{k aces in second deck}]$

I leave it to you to work out that

$P[\text{pick an ace originally in second deck}] = \displaystyle \sum_{k=0}^3~ \dfrac{k}{27} \dfrac{ \dbinom{3}{k} \dbinom{48}{26-k}}{ \dbinom{51}{26}}$

so that

$P[\text{pick ace from 2nd deck}] = \dfrac {1}{27} + \displaystyle \sum_{k=0}^3~ \dfrac{k}{27} \dfrac{ \dbinom{3}{k} \dbinom{48}{26-k}}{ \dbinom{51}{26}}$

3. ## Re: Probability question

But u didn't use conditional probability which is required by the question.

4. ## Re: Probability question

Originally Posted by elmomleo
But u didn't use conditional probability which is required by the question.
I did, it's just hidden.

\begin{align*} &P[\text{picking ace from second deck}] = \\ &P[\text{picking ace from second deck | pick the ace from first deck}]P[\text{pick ace from first deck}] + P[\text{picking ace from second deck | pick an ace originally in second deck}]P[\text{pick ace originally from second deck}] = \\ &1 \cdot P[\text{pick ace from first deck}] + 1 \cdot P[\text{pick ace originally from second deck}] \end{align*}