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6 cards

Wilmer

In Memoriam
Mar 19, 2012
376
Code:
1:                                  1 (26/52)*
2:           11 (12/51)**                                     12 (13/51)***
3:  111            112                          121                        122
4:  1112     1121        1122           1211         1212            1221       1222
5:  11122    11212   11221  11222       12112    12121  12122    12211  12212   12221
6:  111222   112122  112212 112221      121122   121212 121221   122112 122121  122211
6 cards are drawn at random (no replacement) from a regular 52 cards deck.

What is probability that 3 will be hearts and 3 will be diamonds?

1 = hearts : 2 = diamonds
* 26/52 chance to draw 1 or 2 at 1st draw: make it a 1.
** 12/51 chance that 2nd card is a 1
*** 13/51 chance that 2nd card is a 2
...and so on

Is that the proper way to "attack" this?

My 1st time dealing with "probability trees"; doesn't seem like much fun!!

I got probability of 1/250 by simulation...trying to confirm it...
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
I'm not completely confident in this approach but I believe the answer can be found simply by:

\(\displaystyle \frac{ \binom{13}{3} \binom{13}{3}}{\binom{52}{6}}\)

This gives a different answer than you found but this is normally how I approach card problems with no replacement.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Thanks; that computes to .00401777732...
Which is slightly above what I got: 1/250 = .004
So at least my simulation is confirmed (Nod)