Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 05 May 2016, 07:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 17 Nov 2010
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 0

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

### Show Tags

08 Dec 2010, 19:20
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (02:30) wrong based on 3 sessions

### HideShow timer Statictics

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
Math Expert
Joined: 02 Sep 2009
Posts: 32641
Followers: 5655

Kudos [?]: 68706 [0], given: 9817

Re: Confused with two different solutions [#permalink]

### Show Tags

09 Dec 2010, 02:20
Expert's post
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.
_________________
Intern
Joined: 17 Nov 2010
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Confused with two different solutions [#permalink]

### Show Tags

09 Dec 2010, 02:52
Whoah! I understand it now. Can I borrow your brain for my GMAT? =b
Re: Confused with two different solutions   [#permalink] 09 Dec 2010, 02:52
Similar topics Replies Last post
Similar
Topics:
A box contains 5 red & 6 yellow balls. If 4 balls are chosen at random 5 13 Dec 2014, 01:47
1 There are 3 black balls and 10 white balls. If one is to pick 5 balls, 4 21 Nov 2011, 11:27
2 A bag contains 3 white balls, 3 black balls & 2 red balls. 13 10 Aug 2010, 17:41
26 A box contains 3 yellow balls and 5 black balls. One by one 17 09 Feb 2010, 22:00
34 A box contains 3 yellow balls and 5 black balls. One by one 24 29 Sep 2009, 01:14
Display posts from previous: Sort by