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You select 3 cards without replacement from a deck of 52 cards. Calculate the probability that the first card picked was an ace conditional on the fact that the last two were Jacks. Note that here you need to assume that you are drawing the cards out sequentially .

I think I need to do: P(Ace|Jacks) = (P(Ace ∩ Jacks)) / P(Jacks) But I have hard time to find (P(Ace ∩ Jacks)) and P(Jacks). Can anyone explain this for me? Thanks!

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$P(\text{Ace}\cap\text{Jacks})=\frac{(4)\cdot(4\cdot 3)}{52\cdot 51\cdot 50}$

For the first draw you have 4 aces, and the second and the third you have 4 respectively 3 jacks to draw from

$P(\text{Jacks})=\frac{4\cdot 3\cdot 2+48\cdot 4\cdot 3}{52\cdot 51\cdot 50}$

If the first one is jack (4 cases), then for the second and third you need to draw from 3 and 2 jacks

If the first one is not jack (48 cases) you draw from 4 and 3 jacks.

Note that you don't need the denominator (it simplifies).

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An alternate approach:

Reason that the probability that first is ace given second two are jacks is the same as the probability that the third is an ace given the first two are jacks.

Further reason that the probability is the same as the probability of drawing an ace from a deck of 50 cards, four of which are aces.

The probability is then $\frac{4}{50}=\frac{2}{25}=0.08$

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$P($Ace $\cap$ Jacks) is simply the probability of drawing the first card ace and the last two jacks. That should be: $\frac{4}{52} \times \frac{4}{51} \times \frac{3}{50}$.

To find $P($Jacks) you need to consider the two cases, when the first one is jack and when it's not. Therefore the $P($Jacks) $= \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} + \frac{48}{52} \times \frac{4}{51} \times \frac{3}{50}$

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  • $\begingroup$ I got this part. How about P(Jacks)? Should this be (4/51)*(3/50) $\endgroup$ – user6428015 Jan 29 '17 at 23:23
  • $\begingroup$ @user6428015 I will add that part to my answer too $\endgroup$ – Stefan4024 Jan 29 '17 at 23:23

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