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A box contains 1 blue ball, 1 yellow ball,

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A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1

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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 10 Jan 2017, 13:54
Probability that 2nd ball is yellow:
P of 1st ball not Y * P of 2nd ball Y
3/4*1/3= 1/4

P of 3rd ball Not Y:
=1- p of 3rd ball Y
=1- P of 1st ball not Y*p of 2nd ball Not Y* P of 3rd ball Y
=1- 3/4*2/3*1/2
=1-1/4
=/4

So 2nd ball is yellow or the 3rd ball is NOT yellow= 1/4+3/4= 1
What did I do wrong?
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A box contains 1 blue ball, 1 yellow ball, [#permalink]

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GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


Previous question was great and this one is even better. Thanks fo posting!

Approach #1 (Using compliment)

We'll find the probability of a compliment of our statement, that is 2nd ball is not yellow and 3d is yellow which is identical to the statement that 3d is yellow because we have only one yellow ball. We have 3 cases:

Blue Red Yellow + Red Blue Yellow + Red Red Yellow

\(\frac{1}{4}*\frac{2}{3}*\frac{1}{2} + \frac{2}{4}*\frac{1}{3}*\frac{1}{2} + \frac{2}{4}*\frac{1}{3}*\frac{1}{2} = \frac{3}{12} = \frac{1}{4}\)

Our sought-for probability will be \(1 - \frac{1}{4} = \frac{3}{4}\)

Answer C B

Approach #2 (Combinatorics)

"The second ball is yellow" means that we can't choose balls for the second draw, but we can choose balls for the first and the third draw. We need to chose 2 balls from remaining 3 (with exception of yellow) 3C2. Total number of cases will be 4C3.

\(\frac{3C2}{4C3} = \frac{3}{4}\)

Answer C. B

Originally posted by vitaliyGMAT on 10 Jan 2017, 13:59.
Last edited by vitaliyGMAT on 12 Jan 2017, 06:55, edited 1 time in total.
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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nahid78 wrote:
Probability that 2nd ball is yellow:
P of 1st ball not Y * P of 2nd ball Y
3/4*1/3= 1/4

P of 3rd ball Not Y:
=1- p of 3rd ball Y
=1- P of 1st ball not Y*p of 2nd ball Not Y* P of 3rd ball Y
=1- 3/4*2/3*1/2
=1-1/4
=/4

So 2nd ball is yellow or the 3rd ball is NOT yellow= 1/4+3/4= 1
What did I do wrong?


HINT: P(A or B) = P(A) + P(B) - P(A and B)

Cheers,
Brent
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A box contains 1 blue ball, 1 yellow ball, [#permalink]

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GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


Another solution is to list all of the possible outcomes for the 3 selections:
- RRY
- RRB
- RBR
- RYR
- RYB
- RBY
- BYR
- YBR
- BRY
- YRB
- BRR
- YRR

Of the 12 possible outcomes, 9 meet one or both of requirements.
So, P(2nd ball is yellow or the 3rd ball is NOT yellow) = 9/12 = 3/4 = B
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Originally posted by GMATPrepNow on 10 Jan 2017, 14:44.
Last edited by GMATPrepNow on 11 Jan 2017, 08:23, edited 1 time in total.
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


First of all, it's useful to recognize that P(2nd ball is yellow or the 3rd ball is NOT yellow) is the SAME as P(1st ball is yellow or the 2nd ball is NOT yellow)

Let's apply the OR probability rule: P(A or B) = P(A) + P(B) - P(A and B)
So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)

P(1st ball is yellow) = 1/4

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4

P(1st ball is yellow AND the 2nd ball is NOT yellow) = P(1st ball is yellow) x P(the 2nd ball is NOT yellow)
= 1/4 x 1
= 1/4

So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)
= 1/4 + 3/4 - 1/4
= 3/4
= B

Cheers,
Brent
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A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 12 Jan 2017, 03:50
GMATPrepNow wrote:
GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


First of all, it's useful to recognize that P(2nd ball is yellow or the 3rd ball is NOT yellow) is the SAME as P(1st ball is yellow or the 2nd ball is NOT yellow)

Hi Brent, can you please help me understand highlighted statement. Thanks in advance!
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 12 Jan 2017, 04:39
GMATPrepNow wrote:
GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


First of all, it's useful to recognize that P(2nd ball is yellow or the 3rd ball is NOT yellow) is the SAME as P(1st ball is yellow or the 2nd ball is NOT yellow)

Let's apply the OR probability rule: P(A or B) = P(A) + P(B) - P(A and B)
So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)

P(1st ball is yellow) = 1/4

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4

P(1st ball is yellow AND the 2nd ball is NOT yellow) = P(1st ball is yellow) x P(the 2nd ball is NOT yellow)
= 1/4 [b]x1
= 1/4

So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)
= 1/4 + 3/4 - 1/4
= 3/4
= B

Cheers,
Brent



Brent, please explain how you come up with P(the 2nd ball is NOT yellow)=1 in that step? as earlier it was =3/4..... it is confusing....
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 12 Jan 2017, 05:00
GMATPrepNow wrote:
nahid78 wrote:
Probability that 2nd ball is yellow:
P of 1st ball not Y * P of 2nd ball Y
3/4*1/3= 1/4

P of 3rd ball Not Y:
=1- p of 3rd ball Y
=1- P of 1st ball not Y*p of 2nd ball Not Y* P of 3rd ball Y
=1- 3/4*2/3*1/2
=1-1/4
=/4

So 2nd ball is yellow or the 3rd ball is NOT yellow= 1/4+3/4= 1
What did I do wrong?


HINT: P(A or B) = P(A) + P(B) - P(A and B)

Cheers,
Brent


I understand this solution. But Is this only applicable in overlapping situation/s?
Sorry if I am talking foolish. P&C and probability.... uff!

Thanks very much, Brent. :)
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1



Hi

You may have been asked two different things -
(a) 2nd is yellow or (b)3rd is not yellow-,
You should recognise that (a) is part of (b) so just find (b)..

3rd can be..
1) blue
The rest three are 2 red and 1 yellow so ways =3!/2!=3..
2) red
The rest three are 1 red, 1 blue and 1 yellow so ways =3!=6..
Total ways=3+6=9..

Overall ways=4!/2!=12..

Probability=9/12=3/4

B
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A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 13 Jan 2017, 20:12
GMATPrepNow wrote:
GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


First of all, it's useful to recognize that P(2nd ball is yellow or the 3rd ball is NOT yellow) is the SAME as P(1st ball is yellow or the 2nd ball is NOT yellow)

Let's apply the OR probability rule: P(A or B) = P(A) + P(B) - P(A and B)
So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)

P(1st ball is yellow) = 1/4

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4

P(1st ball is yellow AND the 2nd ball is NOT yellow) = P(1st ball is yellow) x P(the 2nd ball is NOT yellow)
= 1/4 x 1
= 1/4

So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)
= 1/4 + 3/4 - 1/4
= 3/4
= B

Cheers,
Brent


GMATPrepNow, Could you briefly explain why P(the 2nd ball is NOT yellow)=1?

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4 , the question mention about WITHOUT REPLACEMENT. I am confused about this statement.
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 14 Jan 2017, 09:26
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Top Contributor
ziyuenlau wrote:
GMATPrepNow wrote:
GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


First of all, it's useful to recognize that P(2nd ball is yellow or the 3rd ball is NOT yellow) is the SAME as P(1st ball is yellow or the 2nd ball is NOT yellow)

Let's apply the OR probability rule: P(A or B) = P(A) + P(B) - P(A and B)
So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)

P(1st ball is yellow) = 1/4

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4

P(1st ball is yellow AND the 2nd ball is NOT yellow) = P(1st ball is yellow) x P(the 2nd ball is NOT yellow)
= 1/4 x 1
= 1/4

So, P(1st ball is yellow or the 2nd ball is NOT yellow) = P(1st ball is yellow) + P(2nd ball is NOT yellow) - P(1st ball is yellow AND the 2nd ball is NOT yellow)
= 1/4 + 3/4 - 1/4
= 3/4
= B

Cheers,
Brent


GMATPrepNow, Could you briefly explain why P(the 2nd ball is NOT yellow)=1?

P(2nd ball is NOT yellow) = P(1st ball is NOT yellow) = 3/4 , the question mention about WITHOUT REPLACEMENT. I am confused about this statement.


Good question.
For AND probabilities, the formula is P(A and B) = P(A) x P(B|A), where "P(B|A)" stands for "the probability that event A occurs GIVEN THAT event B occurs"
So, P(1st ball is yellow AND the 2nd ball is NOT yellow) = P(1st ball is yellow) x P(2nd ball is NOT yellow given that the 1st ball is yellow)
So, if the 1st ball is yellow, then the probability is 1 that the 2nd ball is NOT yellow (since there's only 1 yellow ball).

Does that help?

Cheers,
Brent
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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\(1B\)
\(1Y\)
\(2R\)

the questions asks ==> Prob (2nd yellow) OR Prob (1st yellow ,or 2nd yellow, or no yellow at all)

Notice that the prob of 2nd yellow is counted twice, which means we can remove one of them. which leave us with

\(Prob (1st yellow + 2nd yellow+ No yellow)=1-(3rd yellow)\)

\((Y)*(RorB)*(RorB)+(RorB)*(Y)*(RorB)+(RorB)*(RorB)*(RorB)=1/4*3/3*2/2+3/4*1/3*2/2+3/4*2/3*1/2=3/4\)

same thing as:

\(1-(3rd yellow)=1-[(RorB)*(RorB)*(Y)]=1-(3/4*2/3*1/2)=1-1/4=3/4\)
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A box contains 1 blue ball, 1 yellow ball, [#permalink]

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Great Question.

One way to make both the 'listing all the options' and the 'sequence of events' approaches easier is to just think of the balls as belonging to only two types -- Yellow (Y) and Not Yellow (N).

Listing All The Options

If we include a fourth slot to represent the ball not chosen, I think it's easier to see that there are only four (equally likely) outcomes:

YNN N
NYN N
NNY N
NNN Y

Each of these sequences is equally likely (there's nothing that suggests that the yellow is any more likely to show up in any particular position), so we can just notice that in three of the options (1, 2, and 4 above) the yellow is either second OR not third. So there is a 3/4 chance we'll get one of the three permissible options.


Sequence of Events - Direct

There are only three sequences that have either the second ball yellow or have the third ball as not-yellow:

NYN
YNN
NNN

Plugging in numbers, we get:

NYN ----> (3/4)(1/3)(2/2) = 1/4
YNN ----> (1/4)(3/3)(2/2) = 1/4
NNN ----> (3/4)(2/3)(1/2) = 1/4

So the probability of getting one OR the other OR the other sequence = 1/4 + 1/4 + 1/4 = 1/2


Sequence of Events -- Indirect

There is only one way to get NEITHER of the scenarios we want. That's when the third ball IS yellow (otherwise it will definitely be the case that the third ball is not yellow (or the second is)):

NNY ---> (3/4)(2/3)(1/2) = 1/4

So the probability of NOT getting what we want is 1/4. Therefore the probability of getting what we want is 1 - 1/4 = 3/4
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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GMATPrepNow wrote:
A box contains 1 blue ball, 1 yellow ball, and 2 red balls.
Three balls are randomly selected (one after the other) without replacement.
What is the probability that the 2nd ball is yellow or the 3rd ball is NOT yellow?

A) 1/2
B) 3/4
C) 5/6
D) 11/12
E) 1


We are given that a box contains 1 blue ball, 1 yellow ball, and 2 red balls. We need to determine, when selecting 3 balls from the box, the probability that the second ball is yellow OR the third ball is not yellow.

Let’s look at the requirements in two separate scenarios:

Scenario 1: The second ball is yellow

The probability of not choosing the yellow ball in the first draw is 3/4, and the probability of choosing the yellow ball from the remaining 3 balls in the second draw is 1/3 (we do not care which ball is selected in the third draw in this scenario). Therefore, the probability of choosing the yellow ball in the second draw is:

3/4 x 1/3 = 1/4

Scenario 2: The third ball is not yellow. There are several ways this can happen, as follows:

a. The first ball is yellow. If this happens, we know that the third ball cannot be yellow because there is only one yellow ball in the box.

Since there are four balls, the probability of choosing the yellow ball in the first draw is 1/4 (we do not care which balls are selected in the second or third draws in this scenario).

b. The second ball is yellow. If this happens, we know that the third ball cannot be yellow because there is only one yellow ball in the box. We already considered this outcome in Scenario 1, so we WILL NOT include it again here.

c. The yellow ball is not selected at all. This outcome also satisfies the requirement that the third ball is not yellow.

The probability of choosing a non-yellow ball in the first, second, and third draws is 3/4, 2/3, and 1/2, respectively. So the probability of not choosing a yellow ball in three draws is:

3/4 x 2/3 x 1/2 = 1/4

Since any of these outcomes would satisfy the given condition and they are mutually exclusive, we can sum the individual probabilities of each outcome:

1/4 + 1/4 + 1/4 = 3/4

Answer: B
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Re: A box contains 1 blue ball, 1 yellow ball, [#permalink]

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New post 29 Jan 2018, 04:48
P(2nd yellow) = first not yellow * second yellow = (3/4) * (1/3) = 1/4

P(3rd not yellow) = first yellow + second yellow + no yellow at all
= 1/4 + 1/4 + (3/4)* (2/3) * (1/2) = 3/4

P(2nd yellow AND 3rd not yellow) = P(choosing yellow in second draw) = 1/4

P(2nd yellow OR 3rd not yellow) = P(2nd yellow) + P(3rd not yellow) - P(2nd yellow AND 3rd not yellow)
= (1/4) + (3/4) - (1/4) = 3/4
Re: A box contains 1 blue ball, 1 yellow ball,   [#permalink] 29 Jan 2018, 04:48
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