# Without replacement

• November 8th 2007, 03:21 AM
DivideBy0
Without replacement
Just checking, thanks.

7 balls are drawn from a box containing 6 blue balls and 3 red balls. Find the probability that only 1 of these is red.

Is that:

$=\frac{{6 \choose 6}{3 \choose 1}}{{9 \choose 7}}=\frac{1}{12}$

There is a box containing 3 red balls, 5 green balls, and 9 blue balls. 10 Balls are drawn without replacement. Find the probability that of these, 2 are red and 3 green.

Is that:

$=\frac{{9 \choose 5}{5 \choose 3}{3 \choose 2}}{{17 \choose 10}}=\frac{945}{4862}$
• November 8th 2007, 07:17 AM
ThePerfectHacker
That looks right. (Clapping)