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Alex75PAris
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Guess Strategy ( if time is less for computation )

Consider there were 5 yellow and 5 blue balls , the probality is 50% for each ball getting selected , hence the answer will somewhere hover around 1/2 , D is closer to 1/2 hence Guess D and move on.
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Bunuel
Alex75PAris
Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)

PROBABILITY:

\(\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

COMBINATORICS:

\(\frac{C^2_4*C^2_6}{C^4_{10}}=\frac{3}{7}\).

Answer: D.

Posted from my mobile device

Can you please explain why you multiplied by 4! / (2! * 2!) while calculating by probability method? Thanks.
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Alex75PAris
Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)

the possible combinations of the marbles that meet the condition(exactly 2 yellow and 2 blue) will be: yybb,bbyy,ybyb,byby,ybby,byyb. For yybb combination(for example) the probability will be (6/10 * 5/9 * 4/8 * 3/7 =1/14), for six possible combinations, the total probability will be 6 * 1/14 =3/7.
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kk47
Bunuel
Alex75PAris
Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)

PROBABILITY:

\(\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

COMBINATORICS:

\(\frac{C^2_4*C^2_6}{C^4_{10}}=\frac{3}{7}\).

Answer: D.

Posted from my mobile device

Can you please explain why you multiplied by 4! / (2! * 2!) while calculating by probability method? Thanks.

Let's say Event(A) : Y comes out
B : B comes out

AA'BB' set of events.
Each permutation of that set of event has a probability associated with it.

Each set of permutation is a possible way to get to the solution.
So, total no of ways these 4 possibilities can be arranged is 4!. But The event of yellow coming out or blue coming out are identical from events perspective. So divide by 2!2!
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Bunuel

Alex75PAris
Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)
PROBABILITY:

\(\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

COMBINATORICS:

\(\frac{C^2_4*C^2_6}{C^4_{10}}=\frac{3}{7}\).

Answer: D.
­Bunuel could you please explain the proability method, or just tell me what this is called so I can google it.
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Light11

Bunuel

Alex75PAris
Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)
PROBABILITY:

\(\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

COMBINATORICS:

\(\frac{C^2_4*C^2_6}{C^4_{10}}=\frac{3}{7}\).

Answer: D.
­Bunuel could you please explain the proability method, or just tell me what this is called so I can google it.
­
P(2 yellow and 2 blue) \(=\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

We are multiplying by 4!/(2!*2!) because the YYBB scenario can occur in 6 different ways: YYBB, YBYB, YBBY, BYYB, BYBY, and BBYY. This is the number of permutations of the 4 letters YYBB, where 2 Y's and 2 B's are identical, calculated as 4!/(2!*2!).

Hope it's clear.­
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Bunuel

Light11

Bunuel
PROBABILITY:

\(\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

COMBINATORICS:

\(\frac{C^2_4*C^2_6}{C^4_{10}}=\frac{3}{7}\).

Answer: D.
­Bunuel could you please explain the proability method, or just tell me what this is called so I can google it.
­
P(2 yellow and 2 blue) \(=\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

We are multiplying by 4!/(2!*2!) because the YYBB scenario can occur in 6 different ways: YYBB, YBYB, YBBY, BYYB, BYBY, and BBYY. This is the number of permutations of the 4 letters YYBB, where 2 Y's and 2 B's are identical, calculated as 4!/(2!*2!).

Hope it's clear.­
­Yup, Thankyou Bunuel.
I was confused thinking that the probabilty of each permutation will be different, simplifying it helped.­
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Setup: 6 yellow, 4 blue → Pick 4, find P(exactly 2 yellow and 2 blue)


Step 1: Probability of ONE specific sequence (YYBB)

Pick 1 (Y): 6/10
Pick 2 (Y): 5/9
Pick 3 (B): 4/8
Pick 4 (B): 3/7

P(YYBB) = (6/10) × (5/9) × (4/8) × (3/7) = 360/5040 = 1/14

Step 2: Multiply by ALL valid arrangements

The question asks "exactly 2Y and 2B" — order doesn't matter. All these sequences count:

YYBB, YBYB, YBBY, BBYY, BYBY, BYYB = 6 total

Why 6? Formula: 4!/(2! × 2!) = 24/4 = 6

Why does each sequence have the same probability? Take BYBY:
(4/10) × (6/9) × (3/8) × (5/7) = 360/5040 = 1/14
Same numbers, different order — multiplication is commutative.

Step 3: Final answer

P = (1/14) × 6 = 6/14 = 3/7

Answer: D (3/7)

N0BU

­Bunuel could you please explain the proability method, or just tell me what this is called so I can google it.
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