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# A bag has 15 marbles , 7 black, 8 red . You pick 5 marbles

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CEO
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A bag has 15 marbles , 7 black, 8 red . You pick 5 marbles [#permalink]

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08 Oct 2003, 21:10
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# 1.
A bag has 15 marbles , 7 black, 8 red . You pick 5 marbles one by one. The first four marbles are red.

1. Whats the probability that the fifth marble is red.
2. Whats the probability that the fifth marble is black.

# 2.
A box contains 10 balls ,7 green, 3 yellow . You pick 5 one by one. Out of the first three, 2 green and 1 yellow.

Whats the probability that the remaining balls are a GREEN and a yellow ?

Thanks
Praetorian

Last edited by Praetorian on 08 Oct 2003, 22:40, edited 1 time in total.
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08 Oct 2003, 22:10
#1
1. Whats the probability that the fifth marble is red.
8/15*7/14*6/13*5/12(four are red)*4/11 (red again)

2. Whats the probability that the fifth marble is black.
8/15*7/14*6/13*5/12(four are red)*7/11 (black)

#2
A box contains 10 balls ,7 green, 3 yellow . You pick 5 one by one. Out of the first three, 2 green and 1 yellow.

Whats the probability that the remaining balls are a red and a yellow ?

P=0 there are no red balls in a box
CEO
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08 Oct 2003, 22:44
stolyar wrote:
#1
1. Whats the probability that the fifth marble is red.
8/15*7/14*6/13*5/12(four are red)*4/11 (red again)

2. Whats the probability that the fifth marble is black.
8/15*7/14*6/13*5/12(four are red)*7/11 (black)

#2
A box contains 10 balls ,7 green, 3 yellow . You pick 5 one by one. Out of the first three, 2 green and 1 yellow.

Whats the probability that the remaining balls are a red and a yellow ?

P=0 there are no red balls in a box

Are you sure that the probability DOES NOT change even if we have knowledge of prior events.

This problem is different in that ...we already know that some event has occured..so wont that have an impact on the probability of a future event

We want to find out P( fifth is red / four are red )

i think your solution to # 1 will be correct ONLY when we want to know whats the probability of picking 5 red balls.

again for #2, i have the same doubt.

thanks
praetorian
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08 Oct 2003, 23:17
You mean this approach?

P(5R)=P(4R)*P(R/4R)
5C5/12C5=5C4/12C4*X
X=[5C5/12C5]/[5C4/12C4]=1/8
CEO
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08 Oct 2003, 23:53
stolyar wrote:
You mean this approach?

P(5R)=P(4R)*P(R/4R)
5C5/12C5=5C4/12C4*X
X=[5C5/12C5]/[5C4/12C4]=1/8

I am not quite sure about what you you did there...

I thought of using bayes theorem..but since GMAT does not test on bayes theorem, i was wondering if there was a simpler way out.

i will check bayes theorem and let you know whether 1/8 is correct.

thanks
praetorian
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09 Oct 2003, 00:00
I employed a basic formula for conditional probability.
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Re: PS : Probability (conditional) [#permalink]

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09 Oct 2003, 07:46
praetorian123 wrote:
# 1.
A bag has 15 marbles , 7 black, 8 red . You pick 5 marbles one by one. The first four marbles are red.

1. Whats the probability that the fifth marble is red.
2. Whats the probability that the fifth marble is black.

# 2.
A box contains 10 balls ,7 green, 3 yellow . You pick 5 one by one. Out of the first three, 2 green and 1 yellow.

Whats the probability that the remaining balls are a GREEN and a yellow ?

Thanks
Praetorian

Praet,

#1 - Since it's given that the first 4 are red,
Prob. that 5th is red = 4/11 (We have to find out the probability only for the 5th one to be red and not for all 5 to be red.)
Similarly, Prob. that 5th is black = 7/11.

#2 - I am confused as to what the question means.

Amar.
Re: PS : Probability (conditional)   [#permalink] 09 Oct 2003, 07:46
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