Thread: conditional probabilty

A standard pack contain 52 cards. The card are classified into 3 categories.
1.ace(each worth 14 points)
2. picture cards(each worth 12 points)
3. normal cards(each worth 6 points)

After Darren picks 2 cards withour replacement, Alice draws another two cards without replacement. Given that Darren got 2 points for his 2 cards, calculate the probability that Alice draws an ace and a normal card.

Darren shd have drawn an ace and a pic card to get 26 pts.
I know that this shd be conditonal probability and i need to find the intersection of Darren having ace and pic card and alice having ace and normal card. i dont understand how to get it. and is the probaility of Darren getting 26 pts 4/52 x 12/51 + 12/52 x 4/51 ?

Yes, that's correct.

Now, you just need to get the probability that Alice gets an ace and a normal card, after Darren picked his two cards. The total number of aces is then 3, the number of picture cards is 11 and the number of normal cards is 40, for a total of 50 cards left in the pack.

Multiply the probabilities in the case of Darren and Alice together to get the intersection you're looking for.