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# A box contains 3 yellow balls and 5 black balls. One by one

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A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

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08 Dec 2010, 18:20
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Hi I was wondering if anyonce could explain the difference in solution. Why is #1 approached with the product of probability x favorable outcome while #2 is approached with just getting the probability. What is the difference? Why is there that extra step in #1.???

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

2 of the balls are the same colour?

5x1x4x3 / 5x5x5x5 x 4C2 = 72/125

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

A. 1/4
B. 1/2
C. 1/2
D. 5/8
E. 2/3

7/9 x 6/7 x 5/6 x 5/5 = 5/8
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Re: Confused with two different solutions [#permalink]

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09 Dec 2010, 01:20
gmativy wrote:
Hi I was wondering if anyonce could explain the difference in solution. Why is #1 approached with the product of probability x favorable outcome while #2 is approached with just getting the probability. What is the difference? Why is there that extra step in #1.???

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

2 of the balls are the same colour?

5x1x4x3 / 5x5x5x5 x 4C2 = 72/125

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
1/4
1/2
1/2
5/8
2/3
3/4

7/9 x 6/7 x 5/6 x 5/5 = 5/8

Let me offer another approaches:

1. There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that 2 of the balls are the same colour?

So there are five different balls: {1, 2, 3, 4, 5}

So we need the probability of {XXYZ}, for example {2, 2, 4, 5} in any order: $$P=\frac{C^1_5*C^2_4*}{5^4}*\frac{4!}{2!}=\frac{72}{125}$$;

$$C^1_5$$ - # of ways to choose which ball will be X (the ball which will be chosen twice);
$$C^2_4$$ - # of ways to choose which balls will be Y and Z (the other 2 balls from 4 left);
$$5^4$$ - total # of outcomes;
$$\frac{4!}{2!}$$ - {XXYZ} can occur in several ways: XXYZ, ZYXX, XYXZ, ... basically the # of permutations of 4 letters out of which 2 are identical, which is 4!/2!.

2. A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

There is a shortcut solution for this problem:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Check similar problems:
s99-101553.html?hilit=initial%20probability%20successive#p787694
what-is-the-probability-of-red-ball-100023.html?hilit=initial%20probability%20successive#p771172

Also Probability and Combinatorics chapters of Math Book: math-probability-87244.html and math-combinatorics-87345.html

Hope it helps.
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Re: Confused with two different solutions [#permalink]

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09 Dec 2010, 01:52
Whoah! I understand it now. Can I borrow your brain for my GMAT? =b
Re: Confused with two different solutions   [#permalink] 09 Dec 2010, 01:52
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# A box contains 3 yellow balls and 5 black balls. One by one

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