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A box contains 1 blue ball, 1 green ball, 1 yellow ball

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GreenlightTestPrep
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A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   28 Feb 2018, 12:43
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A box contains 1 blue ball, 1 green 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 NOT red and the 3rd ball is yellow?

Express your answer as a fraction.

Answer:
Show: ::
1/10
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   01 Mar 2018, 15:11
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Since there is only one yellow ball, of course neither the 1st nor the 2nd can be yellow. But there are two scenarios which can result in the 2nd ball being "not red."

In scenario 1, the 1st ball chosen is red, and the 2nd ball is either blue or green. The odds of three events happening should multiplied. So, the odds of the 1st being red are 2/5, the 2nd being either blue or green are 2/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 2/4 x 1/3 = 1/15

In scenario 2, the 1st ball chosen is either blue or green and the 2nd ball is whichever out of blue or green didn't get chosen the first time. So, the odds of the 1st being either blue or green are 2/5, the 2nd being the other one are 1/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 1/4 x 1/3 = 1/30

Adding these two probabilities, 1/15 and 1/30, results in an answer of 1/10.
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   01 Mar 2018, 17:32
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GreenlightTestPrep wrote:
A box contains 1 blue ball, 1 green 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 NOT red and the 3rd ball is yellow?

Express your answer as a fraction.

Answer:
Show: ::
1/10


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

P(1st ball is NOT red AND the 2nd ball is yellow) = P(1st ball is NOT red) x P(the 2nd ball is yellow)
= 2/5 x 1/4
= 1/10

Aside: The first probability, P(1st ball is NOT red), equals 2/5, because there are 5 balls to choose from, and we cannot choose a red ball (because that's not allowed) AND we cannot choose a yellow ball (because, that ball must be available for the next selection). So, of the 5 possible balls to choose from on the first selection, only 2 balls (the blue and green balls) are permissible.

Cheers,
Brent
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   02 Mar 2018, 14:52
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Great question, BTW.

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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   16 Dec 2018, 19:40
2nd ball is NOT red means first ball is red hence the prob of first ball is 2/5

So, the prob of sec ball is 3/4 and prob of 3rd ball is 1/3
Hence final answer is 1/4

I got this answer. I have gone through the explanation above but i am not able to understand it.
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   26 Jan 2019, 12:37
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First of all, it's useful to recognize that P(2nd ball is NOT red and the 3rd ball is yellow) is the SAME as P(1st ball is NOT red and the 2nd ball is yellow)


Would appreciate if you can shed some more light on this conclusion.
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   28 Nov 2019, 07:33
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akingba wrote:
Quote:
First of all, it's useful to recognize that P(2nd ball is NOT red and the 3rd ball is yellow) is the SAME as P(1st ball is NOT red and the 2nd ball is yellow)


Would appreciate if you can shed some more light on this conclusion.


Here's one way to look at it:

Let's say that we have four letters (A, B, C and D) in a bag.
Now let's randomly select one letter at a time until all four letters have been removed.
So, one possible outcome is: C-A-D-B
Another possible outcome is: A-C-B-D
Another possible outcome is: C-B-D-A
etc

The key concept here is that each of the possible outcomes are EQUALLY LIKELY.
For example: P(A-C-B-D) = P(C-B-D-A)
So, P(C is the 2nd letter and B is the 3rd letter) = P(C is the 1st letter and B is the 2nd letter)

Applying the same logic to the original question, we see that P(2nd ball is NOT red and the 3rd ball is yellow) is the SAME as P(1st ball is NOT red and the 2nd ball is yellow)

Does that help?
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   08 Jul 2020, 07:53
GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
A box contains 1 blue ball, 1 green 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 NOT red and the 3rd ball is yellow?

Express your answer as a fraction.

Answer:
Show: ::
1/10


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

P(1st ball is NOT red AND the 2nd ball is yellow) = P(1st ball is NOT red) x P(the 2nd ball is yellow)
= 2/5 x 1/4
= 1/10

Aside: The first probability, P(1st ball is NOT red), equals 2/5, because there are 5 balls to choose from, and we cannot choose a red ball (because that's not allowed) AND we cannot choose a yellow ball (because, that ball must be available for the next selection). So, of the 5 possible balls to choose from on the first selection, only 2 balls (the blue and green balls) are permissible.

Cheers,
Brent


Can you please let me know if my approach to the sum is correct..

[2/5 (P of blue or green) * 1/4 (P of not red or yellow) * 1/3 (P of yellow) ] + [ 2/5 (P of Red) * 2/4 (P of not red or yellow ie P of Blue or green) * 1/3 (P of yellow) ]
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   09 Jul 2020, 08:35
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GreenlightTestPrep wrote:
A box contains 1 blue ball, 1 green 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 NOT red and the 3rd ball is yellow?

Express your answer as a fraction.

Answer:
Show: ::
1/10


can someone point my mistake please

i did the following

the first ball can be anything except yellow, so it has 4 possabilities
second ball either blue or green hence 2
third ball has 1

so: 4x2x1/(5x4x3)= 2/15
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   11 Jul 2020, 05:33
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Please see the explanation provided by Brent.

It is clear and straight


Regards
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   14 Jul 2020, 12:28
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A box contains 1 blue ball, 1 green 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 NOT red and the 3rd ball is yellow?

Step 1: Understanding the question

As three balls are to be selected, we require three positions or slots to be filled ___ AND ____ AND ___

As 3rd ball is yellow and there is only one yellow ball, which is for the third position. Hence, 1 way to fill the 3rd slot.
___ * ____ * 1

2nd ball is not red, it cannot be yellow even, as only one yellow ball is there which is for the third position.
Hence, either blue or green ball can be selected in the 2nd position. There are 2 ways to fill 2nd slot.
___ * 2 * 1

1st ball can either out of 2 red balls or 1 other ball (out of blue or green), hence total ways to fill 3rd position is 3!/2! = 3 ways.
3 * 2 * 1

Step 2: Calculation

Desired outcomes = 3*2*1 = 6

Total outcomes = 5*4*3 = 60 (as balls are picked one after the other, without replacement)

Therefore, probability that the 2nd ball is NOT red and the 3rd ball is yellow is 6/60 = 1/10
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   15 Jul 2020, 01:04
Can you guys recommend me a good book to learn more about this ?
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   15 Jul 2020, 09:54
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Here for FREE

GRE Prep Club - MATH Book for the GRE Exam
https://gre.myprepclub.com/forum/gre-prep- ... html#p5077

And Here a lot of resources

Free GRE Materials - Where to get it!
https://gre.myprepclub.com/forum/free-gmat ... tml#p53911

Regards
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   22 Jul 2021, 06:20
SherpaPrep wrote:
Since there is only one yellow ball, of course neither the 1st nor the 2nd can be yellow. But there are two scenarios which can result in the 2nd ball being "not red."

In scenario 1, the 1st ball chosen is red, and the 2nd ball is either blue or green. The odds of three events happening should multiplied. So, the odds of the 1st being red are 2/5, the 2nd being either blue or green are 2/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 2/4 x 1/3 = 1/15

In scenario 2, the 1st ball chosen is either blue or green and the 2nd ball is whichever out of blue or green didn't get chosen the first time. So, the odds of the 1st being either blue or green are 2/5, the 2nd being the other one are 1/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 1/4 x 1/3 = 1/30

Adding these two probabilities, 1/15 and 1/30, results in an answer of 1/10.



Hey there,

I am really confused about when to add or multiply the probabilities?

I am trying to follow this rule but the step seems different

If all the events happen (an "and question") Multiply the probabilities together.
If only one of the events happens (an "or question") Add the probabilities together

would you shed some light on it?
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   22 Jul 2021, 20:35
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mrk9414 wrote:
SherpaPrep wrote:
Since there is only one yellow ball, of course neither the 1st nor the 2nd can be yellow. But there are two scenarios which can result in the 2nd ball being "not red."

In scenario 1, the 1st ball chosen is red, and the 2nd ball is either blue or green. The odds of three events happening should multiplied. So, the odds of the 1st being red are 2/5, the 2nd being either blue or green are 2/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 2/4 x 1/3 = 1/15

In scenario 2, the 1st ball chosen is either blue or green and the 2nd ball is whichever out of blue or green didn't get chosen the first time. So, the odds of the 1st being either blue or green are 2/5, the 2nd being the other one are 1/4, and the odds of the 3rd being yellow are 1/3:

2/5 x 1/4 x 1/3 = 1/30

Adding these two probabilities, 1/15 and 1/30, results in an answer of 1/10.



Hey there,

I am really confused about when to add or multiply the probabilities?

I am trying to follow this rule but the step seems different

If all the events happen (an "and question") Multiply the probabilities together.
If only one of the events happens (an "or question") Add the probabilities together

would you shed some light on it?


Hi There!

Refer the below:
**And means that the outcome has to satisfy both conditions at the same time.- Multiply
**Or means that the outcome has to satisfy one condition, or the other condition, or both at the same time.- Add

You can also refer the above book for detailed explanation:
Maths book- https://gre.myprepclub.com/forum/gre-math- ... -2609.html

To explain the above question:

We know the third ball is yellow thus, the probability to select a yellow ball= 1/5
The second ball is not red, which means it is either blue or green= 2/4= 1/2
And the first ball is either red or green or blue depending on the second ball= 3/3=1

Now, using the above rule
This question uses the And rule:\(\frac{ 1}{5 }\)x \(\frac{1}{2}\)
=\(\frac{1}{10}\)

Answer:\(\frac{1}{10}\)


Hope this helps!
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   08 Nov 2021, 14:39
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first guest the first ball was red =2/4*1/3=1/5
second option the first ball was not red =1/4*1/3=1/12
the answer is 1/5+1/12=17/60
and this is the only correct answer
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   19 Jul 2022, 18:11
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I took another approach and got lost. I tried listing the valid arrangements and was stuck with 4 until I realized the two red balls created, not two, but four arrangements: BGY, GBY, R1BY, R1GY, R2BY, R3GY: 6 possible arrangements. Total number of arrangements = 5 permutation 3 = 60. Probability = 6/60 = 1/10.
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   04 Apr 2023, 23:44
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possible combinations of the three balls are: rgy, rby, gby, bgy
total combination of the three balls: 5C3= 10

thus 4/10 what am i missing here?
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   05 Apr 2023, 05:34
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abel155 wrote:
possible combinations of the three balls are: rgy, rby, gby, bgy
total combination of the three balls: 5C3= 10

thus 4/10 what am i missing here?


There are a couple of problems with your solution.

First there are more than 4 possible combinations that satisfy the given conditions.
Important: Since there are TWO red balls, we can designate them as R and r, which means our ball options are: b, g, y, R, and r
This means there are 6 possible outcomes: rgy, rby, gby, bgy, Rgy, Rby,

Also, since the order of the selected balls matters, we can't use combinations.
Instead, we must use permutations.
So your calculation should be 5P3 = 60

So our probability = 6/60, which simplifies to be 1/10
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Re: A box contains 1 blue ball, 1 green ball, 1 yellow ball [#permalink] New post   31 Jul 2024, 08:09
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