Question

There are 15 marbles in a bag; 7 of them are red, the other 8 are not red. 4 marbles are drawn WITHOUT replacement.

What is the probability that you'll draw:
1) 0 red
2) 1 red
3) 2 red
4) 3 red
5) 4 red

Thanks in advance for any help :)

Answer 1

0 red: 8/15 * 7/14 * 6/13 * 5/12 = 2/39

1 red: one way would be R, NR, NR, NR (R = red, NR = not red)
= 7/15 * 8/14 * 7/13 * 6/12 =14/195
BUT, the above can be arranged in 4 ways, so prob (1 red) = 56/195

2 red: could be R, R, NR, NR
= 7/15 * 6/14 * 8/13 * 7/12 = 14/195
this can be arranged in 4!/(2!2!) or 6 ways, so prob (2 red) = 28/65

3 red: could be R,R,R,NR
= 7/15 * 6/14 * 5/13 * 8/12 = 2/39
arranged in 4 ways, so prob(3red) = 8/39

4 red: = 7/15 * 6/14 * 5/13 * 4/12 = 1/39



notice: 2/39 + 56/195 + 28/65 + 8/39 + 1/39
= 10/195 + 56/195 + 84/195 + 40/195 + 5/195
= (10 + 56 + 84 + 40 + 5)/195 = 195/195 = 1 , yeahhhh

my answers are correct

Answer 2

Awesome. Thanks so much for your help!

It's funny actually, because I had the same logic, same numbers, but I realized I made a silly calculator error!