# Math Help - simple question about w/o replacement

1. ## simple question about w/o replacement

a box contains 15 balls (4 yellow, 5 purple, 6 blue) 3 balls are drawn at random...

without replacement ..calculate P( all 3 balls are of the same color)

The solution says it should be (4/15)^3 + (5/15)^3 +(6/15)^3

I think it should be (4*3*2)/(15*14*13)...Anyone can explain it to me?

Thanks a lot!

2. ## solved

Ok, so you are close to the answer. Basically, remember that the probability of getting 3 yellows are: $(4 \times 3 \times 2)/(15 \times 14 \times 13)$

3 purples are: $(5 \times 4 \times 3)/(15 \times 14 \times 13)$

3 blue are: $(6 \times 4 \times 3)/(15 \times 14 \times 13)$

Hence, when you combine the terms, you get $((4 \times 3 \times 2) + (5 \times 4 \times 3) + (6 \times 4 \times 3))/(15 \times 14 \times 13)$

hope that clears it up, have fun.