Probability of picking a ball randomly from a group of balls

1. Dec 3, 2012

logearav

1. The problem statement, all variables and given/known data

From a bag containing 4 white and 6 black balls, 2 balls are drawn at random. If the balls are drawn one after the other, without replacement, find the probability that
The first ball is white and the second ball is black
2. Relevant equations

3. The attempt at a solution
First ball is white so the probability is 4/10; Second ball is black so the probability is 6/9. Hence the probability of first ball is white and the second ball is black is given by 4/10 X 6/9 = 24/90 when simplified gives 4/15.
But when i try in the following way
The probability of picking two balls from 10 balls is 10C2 that is 45.
The probability of first ball being white is 4C1 that is 4 and the second ball being black is 6C1 that is 6, so the probability for the question is n(E)/n(S) = (4 X 6)/10C2 = 24/45 which gives 8/15.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 4, 2012

Sourabh N

Since the order (in which balls are picked) matters,

3. Dec 4, 2012

haruspex

Well, not probability, of course. That's the number of ways of picking an unordered pair of balls.
That's the number of pairs of balls, one being white and the other black. So 8/15 is the probability of getting one of each, in either order.

4. Dec 4, 2012

Ray Vickson

Since the balls are distinguishable (in principle you could paint identifying numbers 1--10 on them) and order matters you should look at *permutations*. There are 10! distinct permutations, but we just look at the first two places. There are 4*6*8! different permutations in which the first place is occupied by a white ball and the second place by a black ball, so the probability is 4*6*8!/10! = 4/15, as you stated originally. Your second way ignores the order.