Probability of picking a ball randomly from a group of balls

[logearav]

Homework Statement

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

Homework Equations

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.
Please help me where i went wrong.

 

[Sourabh N]

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

The probability of picking two balls from 10 balls is ¹⁰C₂ that is 45.

should instead be ¹⁰P₂.

 

[haruspex]

The probability of picking two balls from 10 balls is ¹⁰C₂ that is 45.

Well, not probability, of course. That's the number of ways of picking an unordered pair of balls.

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.

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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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.
Please help me where i went wrong.

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.