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# You have a deck of 60 cards: 15 Red, 45 Blue. You draw 6 cards. What are the odds that you draw 2 Red?

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You have a deck of 60 cards: 15 Red, 45 Blue. You draw 6 cards. What are the odds that you draw 2 Red?

Guest Mar 28, 2015

#2
+80823
+10

The number of possible hands of drawing any 6 cards from 60 = C(60,6) = 50063860

And from these possibilities we want to draw  2 red cards from 15 and 4 blue cards from 45

So, this gives us

C(15.2) * C(45, 4) = 15644475

So

15644475 / 50063860   = anout 31.25%

CPhill  Mar 28, 2015
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#1
+17705
+10

To get red-red-blue-blue-blue-blue would be 15/60 x 14/59 x 45/58 x 44/57 x 43/56 x 42/55.

But this is not the only arrangement; there are actually 15 ways to end with two reds and four blues, so take the above answer times 15.

geno3141  Mar 28, 2015
#2
+80823
+10

The number of possible hands of drawing any 6 cards from 60 = C(60,6) = 50063860

And from these possibilities we want to draw  2 red cards from 15 and 4 blue cards from 45

So, this gives us

C(15.2) * C(45, 4) = 15644475

So

15644475 / 50063860   = anout 31.25%

CPhill  Mar 28, 2015
#3
+91412
+5

As you can see it is the same as CPhill's answer.    :)

$$\left({\frac{{\mathtt{15}}}{{\mathtt{60}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{14}}}{{\mathtt{59}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{45}}}{{\mathtt{58}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{44}}}{{\mathtt{57}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{43}}}{{\mathtt{56}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{42}}}{{\mathtt{55}}}}\right){\mathtt{\,\times\,}}{\mathtt{15}} = {\mathtt{0.312\: \!490\: \!387\: \!277\: \!369\: \!3}}$$

Melody  Mar 28, 2015

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