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Certain bowl contains 5 red marbles and 3 blue marbles only.

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by [email protected] » Sun Apr 07, 2013 11:40 pm
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
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Source: — Problem Solving |

by Anju@Gurome » Mon Apr 08, 2013 12:00 am
[email protected] wrote:Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
The problem basically asks the probability that the seventh marble is a red marble.
Now, there are 5 red marbles among a total of 8 marbles.
Hence, the probability of seventh marble being red = 5/8
Last edited by Anju@Gurome on Mon Apr 08, 2013 12:04 am, edited 1 time in total.
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by Anju@Gurome » Mon Apr 08, 2013 12:01 am
Another approach can be as follows...

The problem can be translated to "If 5 red marbles and 3 blue marbles are arranged in a row, what is the probability that the last marble is red?"

Total number of arrangements = 8!/[(5!)*(3!)] = 8*7*6/3! = 8*7
Number of arrangements such that last marble is red = Number of arrangements for 4 red marbles and 3 blue marbles = 7!/[(4!)*(3!)] = 786*5/3! = 7*5

Hence, required probability = 7*5/(887) = 5/8
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by GMATGuruNY » Mon Apr 08, 2013 12:55 am
[email protected] wrote:Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
In other words, what is the probability that the seventh marble selected is red?

No math is needed here if we understand the following concept:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(red on the 7th pick) = P(red on the first pick) = 5/8.

The answer would be the same even if the problem were as follows:
A certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the FIFTH marble drawn is NOT blue?
P(red on the 5th pick) = P(red on the 1st pick) = 5/8.

Other problems that test the same concept:

https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html
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by vipulgoyal » Mon Apr 08, 2013 1:06 am
P 7th marble is not blue =

1 - P(7th M blue)

1-3/8 = 5/8
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by rac.nishu » Mon Jul 01, 2013 5:08 am
The fact that marbles are drawn without replacement does not change anything? I have my doubts on this one .. can anyone please explain!
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