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Bunuel
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Math: Probability 16 Aug 2022, 00:39
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Blair15
Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?

Can someone pls explain how to solve this question? I didn't get the probability tree approach.
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Check here: https://gmatclub.com/forum/julia-and-br ... 98590.html Hope it helps.
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Moaz1996
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Math: Probability 06 Dec 2022, 21:16
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Quote:
Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?
Solution:

1) combinatorial approach: The total number of possible committees is N=C82N=C28. The number of possible committee that includes both Bob and Rachel is n=1n=1.
P=nN=1C82=128P=nN=1C28=128

2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is N=C82N=C28. The number of possible committee that does not includes both Bob and Rachel is:
m=C62+2∗C61m=C26+2∗C16 where,
C62C26 - the number of committees formed from 6 other people.
2∗C612∗C16 - the number of committees formed from Rob or Rachel and one out of 6 other people.
P=1−mN=1−C62+2∗C61C82P=1−mN=1−C26+2∗C16C28
P=1−15+2∗628=1−2728=128
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Hello Walker, thanks for the great post!

I don't understand the second part of (2) and was hoping you could help answer a specific question.

For the number of committees formed without one of Bob or Rachel, I'm confused about the value of n in nCr. You have used n = 6 (6C1*2). Shouldn't we instead use 7C1*2, where 7 represents the total number of people without Bob or Rachel and we pick out 2?
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Soumya531
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Math: Probability 19 Aug 2025, 22:07
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For the equation:
X + Y + Z = 10 where each X, Y and Z integers and are >=0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


The solution is 12C2 and not 14C4.
abagnale11
Let me briefly talk about a concept that i remember studying long back.Most of the logic would be excerpts from Higher Algebra by Bernard Shaw

We know that most of probability questions can be easily cracked if we have a strong grip on Permutation and combination.Let me present you all with something interesting. I'll be dealing with "Distribution" of similar and dissimilar things

Concept 1: Distributing Similar things among people where each recipient receives at least 1 thing
----------------------------------------------------------------------------------------------

Imagine you have 10 similar tennis balls which you need distribute it to 3 friends and each of the fella receives at least 1 ball.

how would you approach it???

Let us do it figuratively.Imagine the 10 balls

O O O O O O O O O O ( remember there are 9 spaces between 10 balls)

if you consider 2 sticks ! !, how many ways can u place them in the 9 spaces between the 10 balls. yeah right 9C2 ways.. would u believe , thats the answer. for example consider O O ! O O O ! O O O O O . this is one of the 9C2 ways where the first guy gets 2 balls, 2 nd guy gets 3 and the last gets 5. you just need to place (r-1) sticks in the spaces between the n balls and that's like selecting r-1 spaces between the n balls.
so the answer is n-1Cr-1 n= number of balls r= number of recepients
I can rephrase the same question

Find the number of solution to the equation
X + Y + Z = 10 where each X, Y and Z integers and are >0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


hence u already know the solution 9C 2 :shock:


Concept 2 : Distributing Similar things among people where each recipient may receive no balls at all
----------------------------------------------------------------------------------------------

Imagine the same old story where you have 10 similar tennis balls which you need distribute it to 5 friends. However now 1 or more of them may receive 0 balls

Again imagine the balls lying beautifully in the imagination of your brain

O O O O O O O O O O .

Now as you know for 5 friends you need to consider 4 sticks ! ! ! !. Now that any of the friend may receive 0 balls also, we cannot consider the space between the balls. So what do you think.. let me figuratively sample u some solution

one of the solution O ! O ! O O ! O O O ! O O O ( FRIEND 1= 1 BALL, FRIEND 2= 1 BALL FRIEND 3= 2 BALLS FRIEND 4= 3 BALLS FRIEND 5 = 3BALLS)
ANOTHER SOLUTION ! O O O ! O O O ! O O O O ! ( FRIEND 1 = 0 BALLS , FRIEND 2= 3 BALLS , FRIEND 3= 3 BALLS , FRIEND 4 = 4 BALLS FRIEND 5 = 0)

I hope by now you would have guessed the solution is nothing but arranging 14 things of which 10 are of one kind and 4 are of one kind

that is !14/(!10*!4) . this is nothing but n+r-1 C n where n is the number of balls and r is the number of recipients

I can again rephrase the same question

Find the number of solution to the equation
X + Y + Z = 10 where each X, Y and Z integers and are >=0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


hence u already know the solution 14C4 :shock:



I can present some more topics. this will be based on the condition that the readers find it interesting and helpful
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