Question

From a standard deck of 52 cards we randomly select seven without replacement.
What is the probability that the 2nd, 3rd and 4th cards are hearts, given that the
1st, 5th, 6th, and 7th cards are hearts?

Answer 1

Okay, let's start with a simple definition we have: \[
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}
\]

Answer 2

We will say \(A\) is the event that 2,3,4 are hearts. \(B\) is the event that 1,5,6,7 are hearts.

Answer 3

Can you compute: \(\Pr(A\cap B)\)?

Answer 4

is it 13/52 *12/51* 11/50* 10/49 *9/48* 8/47* 7/47 ?

Answer 5

yes

Answer 6

Does p(B)= (13/52* 9/48* 8/47* 7/46) + ( 13/52 *12/48 *11/47 *10/46 ) ?

Answer 7

Do you know how to get \(\Pr(B)\)?

Answer 8

Getting the probability of \(B\) is a bit tough.

Answer 9

does my answer right?

Answer 10

No

Answer 11

could you help me with that, please?

Answer 12

Okay.

Answer 13

Thanks

Answer 14

First, you have \(13/52\), for the first card.

Answer 15

For cards 2,3,4 you don't know what will happen, so that much part is tough.

Answer 16

So you don't know if you'll have 12 cards remaining, or you'll have 9 cards remaining.

Answer 17

12

Answer 18

You have 4 cases...
Case 1: none of 2,3,4 are hearts
Case 2: one of 2,3,4 is heart
Case 3: two of 2,3,4 are hearts
Case 4: all of 2,3,4 are hearts

Answer 19

For the first case... There are \(
{3 \choose 0}
\) ways for it to happen.

Answer 20

So the probability of \(B\) is given by: \[
\left(\frac{13}{54}\right)\sum_{k=0}^3{3\choose k}\left(\frac{12-k}{48}\right) \left(\frac{11-k}{47}\right) \left(\frac{10-k}{46}\right)
\]

Answer 21

It doesn't look very nice

Answer 22

13/52 the first card, right

Answer 23

oh yeah.

Answer 24

But you understand why it's complicated right?

Answer 25

yes, i do
Thanks I appreciate you help