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Probability of picking a ball randomly from a group of balls

  • Thread starter logearav
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1. 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
2. Homework 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.
Please help me where i went wrong.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
631
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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

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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

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1. 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
2. Homework 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.
Please help me where i went wrong.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
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.
 

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