Could experts please answer my question below?

GMATGuruNY wrote: Here are all the independent probabilities that would get us a good outcome (black on the 4th pick):

P(BBBB) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

P(YBBB) = 3/8 * 5/7 * 4/6 * 3/5 = 3/28
Since (BYBB) and (BBYB) are also good outcomes, total number of ways to get 1 yellow with black on the 4th pick = 3 * 3/28 = 9/28

P(YYBB) = 3/8 * 2/7 * 5/6 * 4/5 = 1/14
Since (BYYB) and (YBYB) are also good outcomes, total number of ways to get 2 yellow with black on the 4th pick = 3 * 1/14 = 3/14

P(YYYB) = 3/8 * 2/7 * 1/6 * 5/5 = 1/56

Since all of the above are good outcomes, we add the fractions:

1/14 + 9/28 + 3/14 + 1/56 = 35/56 = 5/8.

Much easier just to remember the rule I posted above:

P(B) on the nth pick is the same as P(B) on the 1st pick: 5/8.

I think, the theory that p(a) for the first time = p(a) for nth time is applicable only when the number of balls is greater than n where the questions is about nth position of the ball

I'll just get to what I'm trying to say here:
Balls available = 3 red, 2 blue
P(The third ball picked up is blue) i.e. P(**B)
P(**B) = P(RRB) + P(RBB) + P(BRB) = 1/3 * 1/2 * 1/2 + 1/3 * 1/2 * 1/1 + 1/2 * 1/3 * 1/1
= 1/12 + 1/6 + 1/6
= 5/12 -----------------(1)

whereas, P(first ball is blue) = 2/5 -----------------(2)

(1) and (2) are NOT equal in this case since number of blue balls is only 2 and we're talking about a blue ball on 3rd position.

Does it make sense? Thanks in advance, dear expert.

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Adding one more questions here:
If the balls were to be replaced each time one is picked, the probability according to me would be:
P(RRB)+P(RBB)+P(BRB)+P(BBB)
= 1/3 1/3 1/2 + 1/3 1/2 1/2 + 1/2 1/3 1/2 + 1/2 1/2 1/2
= 1/18 + 1/12 + 1/12 + 1/8
= 25/72

Is the above right?

Thanks in advance

divineacclivity wrote:I think, the theory that p(a) for the first time = p(a) for nth time is applicable only when the number of balls is greater than n where the questions is about nth position of the ball

I'll just get to what I'm trying to say here:
Balls available = 3 red, 2 blue
P(The third ball picked up is blue) i.e. P(**B)
P(**B) = P(RRB) + P(RBB) + P(BRB) = 1/3 * 1/2 * 1/2 + 1/3 * 1/2 * 1/1 + 1/2 * 1/3 * 1/1
= 1/12 + 1/6 + 1/6
= 5/12 -----------------(1)

whereas, P(first ball is blue) = 2/5 -----------------(2)

(1) and (2) are NOT equal in this case since number of blue balls is only 2 and we're talking about a blue ball on 3rd position.

Does it make sense? Thanks in advance, dear expert.

The fractions used to calculate P(**B) without replacement are not quite right.
Given 3 red marbles and 2 blue marbles, P(**B) without replacement can be calculated as follows:

P(RRB):
P(R on the 1st pick) = 3/5. (Of the 5 marbles, 3 are red.)
P(R on the 2nd pick) = 2/4. (Of the 4 remaining marbles, 2 are red.)
P(B on the 3rd pick) = 2/3. (Of the 3 remaining marbles, 2 are blue.)
Multiplying the probabilities, we get:
3/5 * 2/4 * 2/3 = 12/60.

P(RBB):
P(R on the 1st pick) = 3/5. (Of the 5 marbles, 3 are red.)
P(B on the 2nd pick) = 2/4. (Of the 4 remaining marbles, 2 are blue.)
P(B on the 3rd pick) = 1/3. (Of the 3 remaining marbles, 1 is blue.)
Multiplying the probabilities, we get:
3/5 * 2/4 * 1/3 = 6/60.

P(BRB):
P(B on the 1st pick) = 2/5. (Of the 5 marbles, 2 are blue.)
P(R on the 2nd pick) = 3/4. (Of the 4 remaining marbles, 3 are red.)
P(B on the 3rd pick) = 1/3. (Of the 3 remaining marbles, 1 is blue.)
Multiplying the probabilities, we get:
2/5 * 3/4 * 1/3 = 6/60.

Since any of cases above will yield B on the 3rd pick, we add the probabilities:
P(B on the 3rd pick) = 12/60 + 6/60 + 6/60 = 24/60 = 2/5.

The result is the same as P(B) on the 1st pick.

Adding one more questions here:
If the balls were to be replaced each time one is picked, the probability according to me would be:
P(RRB)+P(RBB)+P(BRB)+P(BBB)
= 1/3 1/3 1/2 + 1/3 1/2 1/2 + 1/2 1/3 1/2 + 1/2 1/2 1/2
= 1/18 + 1/12 + 1/12 + 1/8
= 25/72

Is the above right?

Thanks in advance

Just as the fractions used to calculate P(**B) without replacement were not quite right, so too are these fractions.
I suggest that, after reading my solution for P(**B) without replacement, you try again to calculate P(**B) with replacement.

GMATGuruNY wrote:

divineacclivity wrote:I think, the theory that p(a) for the first time = p(a) for nth time is applicable only when the number of balls is greater than n where the questions is about nth position of the ball

I'll just get to what I'm trying to say here:
Balls available = 3 red, 2 blue
P(The third ball picked up is blue) i.e. P(**B)
P(**B) = P(RRB) + P(RBB) + P(BRB) = 1/3 * 1/2 * 1/2 + 1/3 * 1/2 * 1/1 + 1/2 * 1/3 * 1/1
= 1/12 + 1/6 + 1/6
= 5/12 -----------------(1)

whereas, P(first ball is blue) = 2/5 -----------------(2)

(1) and (2) are NOT equal in this case since number of blue balls is only 2 and we're talking about a blue ball on 3rd position.

Does it make sense? Thanks in advance, dear expert.

The fractions used to calculate P(**B) without replacement are not quite right.
Given 3 red marbles and 2 blue marbles, P(**B) without replacement can be calculated as follows:

P(RRB):
P(R on the 1st pick) = 3/5. (Of the 5 marbles, 3 are red.)
P(R on the 2nd pick) = 2/4. (Of the 4 remaining marbles, 2 are red.)
P(B on the 3rd pick) = 2/3. (Of the 3 remaining marbles, 2 are blue.)
Multiplying the probabilities, we get:
3/5 * 2/4 * 2/3 = 12/60.

P(RBB):
P(R on the 1st pick) = 3/5. (Of the 5 marbles, 3 are red.)
P(B on the 2nd pick) = 2/4. (Of the 4 remaining marbles, 2 are blue.)
P(B on the 3rd pick) = 1/3. (Of the 3 remaining marbles, 1 is blue.)
Multiplying the probabilities, we get:
3/5 * 2/4 * 1/3 = 6/60.

P(BRB):
P(B on the 1st pick) = 2/5. (Of the 5 marbles, 2 are blue.)
P(R on the 2nd pick) = 3/4. (Of the 4 remaining marbles, 3 are red.)
P(B on the 3rd pick) = 1/3. (Of the 3 remaining marbles, 1 is blue.)
Multiplying the probabilities, we get:
2/5 * 3/4 * 1/3 = 6/60.

Since any of cases above will yield B on the 3rd pick, we add the probabilities:
P(B on the 3rd pick) = 12/60 + 6/60 + 6/60 = 24/60 = 2/5.

The result is same as P(B) on the 1st pick.

Adding one more questions here:
If the balls were to be replaced each time one is picked, the probability according to me would be:
P(RRB)+P(RBB)+P(BRB)+P(BBB)
= 1/3 1/3 1/2 + 1/3 1/2 1/2 + 1/2 1/3 1/2 + 1/2 1/2 1/2
= 1/18 + 1/12 + 1/12 + 1/8
= 25/72

Is the above right?

Thanks in advance

Just as the fractions used to calculate P(**B) without replacement were not quite right, so too are these fractions.
I suggest that, after reading my solution for P(**B) without replacement, you try again to calculate P(**B) with replacement.

I got your point. On second thoughts, yeah what was I thinking. Thank you very much
P(**B) with replacement = P(RRB)+P(RBB)+P(BRB)+P(BBB)
= 3/5 3/5 2/5 + 3/5 2/5 2/5 + 2/5 3/5 2/5 + 2/5 2/5 2/5
= 2/5 again :O
Why's there no different in the probability with and without replacement. If you think about it, 3rd time I pick a marble (if I'm not replacing), there'd be a case when I don't have a blue marble available (since the earlier ones were both blue) whereas this would never be the case if I'm replacing the marbles each time I pick up one.
Would you please elaborate? I'm not able to wrap my head around it.
Thanks so much in advance

GMATGuruNY wrote:

divineacclivity wrote:I think, the theory that p(a) for the first time = p(a) for nth time is applicable only when the number of balls is greater than n where the questions is about nth position of the ball

I'll just get to what I'm trying to say here:
Balls available = 3 red, 2 blue
P(The third ball picked up is blue) i.e. P(**B)
P(**B) = P(RRB) + P(RBB) + P(BRB) = 1/3 * 1/2 * 1/2 + 1/3 * 1/2 * 1/1 + 1/2 * 1/3 * 1/1
= 1/12 + 1/6 + 1/6
= 5/12 -----------------(1)

whereas, P(first ball is blue) = 2/5 -----------------(2)

(1) and (2) are NOT equal in this case since number of blue balls is only 2 and we're talking about a blue ball on 3rd position.

Does it make sense? Thanks in advance, dear expert.

The fractions used to calculate P(**B) without replacement are not quite right.
Given 3 red marbles and 2 blue marbles, P(**B) without replacement can be calculated as follows:

P(RRB):
P(R on the 1st pick) = 3/5. (Of the 5 marbles, 3 are red.)
P(R on the 2nd pick) = 2/4. (Of the 4 remaining marbles, 2 are red.)
P(B on the 3rd pick) = 2/3. (Of the 3 remaining marbles, 2 are blue.)
Multiplying the probabilities, we get:
3/5 * 2/4 * 2/3 = 12/60.

P(RBB):
P(R on the 1st pick) = 3/5. (Of the 5 marbles, 3 are red.)
P(B on the 2nd pick) = 2/4. (Of the 4 remaining marbles, 2 are blue.)
P(B on the 3rd pick) = 1/3. (Of the 3 remaining marbles, 1 is blue.)
Multiplying the probabilities, we get:
3/5 * 2/4 * 1/3 = 6/60.

P(BRB):
P(B on the 1st pick) = 2/5. (Of the 5 marbles, 2 are blue.)
P(R on the 2nd pick) = 3/4. (Of the 4 remaining marbles, 3 are red.)
P(B on the 3rd pick) = 1/3. (Of the 3 remaining marbles, 1 is blue.)
Multiplying the probabilities, we get:
2/5 * 3/4 * 1/3 = 6/60.

Since any of cases above will yield B on the 3rd pick, we add the probabilities:
P(B on the 3rd pick) = 12/60 + 6/60 + 6/60 = 24/60 = 2/5.

The result is same as P(B) on the 1st pick.

Adding one more questions here:
If the balls were to be replaced each time one is picked, the probability according to me would be:
P(RRB)+P(RBB)+P(BRB)+P(BBB)
= 1/3 1/3 1/2 + 1/3 1/2 1/2 + 1/2 1/3 1/2 + 1/2 1/2 1/2
= 1/18 + 1/12 + 1/12 + 1/8
= 25/72

Is the above right?

Thanks in advance

Just as the fractions used to calculate P(**B) without replacement were not quite right, so too are these fractions.
I suggest that, after reading my solution for P(**B) without replacement, you try again to calculate P(**B) with replacement.

I got your point. On second thoughts, yeah what was I thinking. Thank you very much
P(**B) with replacement = P(RRB)+P(RBB)+P(BRB)+P(BBB)
= 3/5 3/5 2/5 + 3/5 2/5 2/5 + 2/5 3/5 2/5 + 2/5 2/5 2/5
= 2/5 again :O
Why's there no different in the probability with and without replacement. If you think about it, 3rd time I pick a marble (if I'm not replacing), there'd be a case when I don't have a blue marble available (since the earlier ones were both blue) whereas this would never be the case if I'm replacing the marbles each time I pick up one.
Would you please elaborate? I'm not able to wrap my head around it.
Thanks so much in advance

nox104 wrote:

P(B) on the nth pick is the same as P(B) on the 1st pick

Can you explain what this rule actually means in words and when is it applicable?

I followed an easier approach and arrived at the same answer.

First, fixing the 4th pick as a black, we have 7 balls (3 Y + 4 B) remaining. Using the ANAGRAMS technique,

P (B on the 4th) = all arrangements of YYYBBBB / all arrangements of YYYBBBBB

Numerator = 7!/3!4! = 35
Denominator = 8!/3!5! = 56

Answer = 35/56 = 5/8

Can we solve it as below . Since denominator is always 8 * 7 * 6 * 5 =1680 I will take it as common.
We pick Black as fourth ball in below scenarios :-
YYYB -3 * 2 * 1 * 5 = 30/1680
YYBB -3 * 2 * 5 * 4 = 120/1680
YBBB -3 * 5 * 4 * 3 = 180/1680
BBBB -5 * 4 * 3 * 2 = 120/1680

Now
YYYB can arranged in 4!/3! = 4 ways so 4 * 30/1680 = 120/1680
YYBB can arranged in 4!/2!2! = 6 ways so 6 * 120/1680 = 720/1680
YBBB can arranged in 4!/3! = 4 ways so 4 * 180/1680 = 720/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

But total again comes as 1680/1680 as 1 . Please tell where I did wrong.

Can we solve it as below . Since denominator is always 8 * 7 * 6 * 5 =1680 I will take it as common.
We pick Black as fourth ball in below scenarios :-
YYYB -3 * 2 * 1 * 5 = 30/1680
YYBB -3 * 2 * 5 * 4 = 120/1680
YBBB -3 * 5 * 4 * 3 = 180/1680
BBBB -5 * 4 * 3 * 2 = 120/1680

Now
YYYB can arranged in 4!/3! = 4 ways so 4 * 30/1680 = 120/1680
YYBB can arranged in 4!/2!2! = 6 ways so 6 * 120/1680 = 720/1680
YBBB can arranged in 4!/3! = 4 ways so 4 * 180/1680 = 720/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

But total again comes as 1680/1680 as 1 . Please tell where I did wrong.

nikhilgmat31 wrote:Can we solve it as below . Since denominator is always 8 * 7 * 6 * 5 =1680 I will take it as common.
We pick Black as fourth ball in below scenarios :-
YYYB -3 * 2 * 1 * 5 = 30/1680
YYBB -3 * 2 * 5 * 4 = 120/1680
YBBB -3 * 5 * 4 * 3 = 180/1680
BBBB -5 * 4 * 3 * 2 = 120/1680

Now
YYYB can arranged in 4!/3! = 4 ways so 4 * 30/1680 = 120/1680
YYBB can arranged in 4!/2!2! = 6 ways so 6 * 120/1680 = 720/1680
YBBB can arranged in 4!/3! = 4 ways so 4 * 180/1680 = 720/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

But total again comes as 1680/1680 as 1 . Please tell where I did wrong.

Now
YYYB can arranged in 3!/3! = 1 way so 1 * 30/1680 = 30/1680
YYBB can arranged in 3!/2!1! = 3 ways so 3 * 120/1680 = 360/1680
YBBB can arranged in 3!/1!2! = 3 ways so 3 * 180/1680 = 540/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

Total = 1050/1680 = 105/168 = 5/8

theCEO wrote:

nikhilgmat31 wrote:Can we solve it as below . Since denominator is always 8 * 7 * 6 * 5 =1680 I will take it as common.
We pick Black as fourth ball in below scenarios :-
YYYB -3 * 2 * 1 * 5 = 30/1680
YYBB -3 * 2 * 5 * 4 = 120/1680
YBBB -3 * 5 * 4 * 3 = 180/1680
BBBB -5 * 4 * 3 * 2 = 120/1680

Now
YYYB can arranged in 4!/3! = 4 ways so 4 * 30/1680 = 120/1680
YYBB can arranged in 4!/2!2! = 6 ways so 6 * 120/1680 = 720/1680
YBBB can arranged in 4!/3! = 4 ways so 4 * 180/1680 = 720/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

But total again comes as 1680/1680 as 1 . Please tell where I did wrong.

Now
YYYB can arranged in 3!/3! = 1 way so 1 * 30/1680 = 30/1680
YYBB can arranged in 3!/2!1! = 3 ways so 3 * 120/1680 = 360/1680
YBBB can arranged in 3!/1!2! = 3 ways so 3 * 180/1680 = 540/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

Total = 1050/1680 = 105/168 = 5/8

It seems you have applied ANAGRAMS in a different manner, Can you please help me to understand it.

nikhilgmat31 wrote:

theCEO wrote:

nikhilgmat31 wrote:Can we solve it as below . Since denominator is always 8 * 7 * 6 * 5 =1680 I will take it as common.
We pick Black as fourth ball in below scenarios :-
YYYB -3 * 2 * 1 * 5 = 30/1680
YYBB -3 * 2 * 5 * 4 = 120/1680
YBBB -3 * 5 * 4 * 3 = 180/1680
BBBB -5 * 4 * 3 * 2 = 120/1680

Now
YYYB can arranged in 4!/3! = 4 ways so 4 * 30/1680 = 120/1680
YYBB can arranged in 4!/2!2! = 6 ways so 6 * 120/1680 = 720/1680
YBBB can arranged in 4!/3! = 4 ways so 4 * 180/1680 = 720/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

But total again comes as 1680/1680 as 1 . Please tell where I did wrong.

Now
YYYB can arranged in 3!/3! = 1 way so 1 * 30/1680 = 30/1680
YYBB can arranged in 3!/2!1! = 3 ways so 3 * 120/1680 = 360/1680
YBBB can arranged in 3!/1!2! = 3 ways so 3 * 180/1680 = 540/1680
BBBB can be arranged in 1 way so 1 * 120/1680 = 120/1680

Total = 1050/1680 = 105/168 = 5/8

It seems you have applied ANAGRAMS in a different manner, Can you please help me to understand it.

In your post you wrote:
YYYB can arranged in 4!/3! = 4 ways so 4 * 30/1680 = 120/1680
With this calculation you are also including the result for YYBY and YBYY etc
The same thing applies for your other possibilities
With this calculation your are counting all possibilites, hence the reason why your answer comes to 1.

The B term un the 4th position should be held constant. We are interested in the number of ways the first, second and third terms can be arranged

Hi CEO,
Perfect explanation. we need to fix the position B at fourth place.