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## Comment on

Same Color## In this question, isn´t it

## That's a perfectly valid

That's a perfectly valid solution - great work!

## Help for this question plz:

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?

## Here's 1 solution:

Here's 1 solution:

We want to determine P(2nd ball is not red AND 3rd ball is yellow)

First we must recognize that there are TWO WAYS to meet the given conditions:

case #1) 1st ball is not red, 2nd ball is not red, 3rd ball is yellow

case #2) 1st ball is red, 2nd ball is not red, 3rd ball is yellow

P(2nd ball is not red AND 3rd ball is yellow) = P(case #1 OR case #2)

= P(case #1) + P(case #2)

Let's examine each case separately.

P(case #1) = P(1st ball is not red AND 2nd ball is not red AND 3rd ball is yellow)

= P(1st ball is not red) x P(2nd ball is not red) x P(3rd ball is yellow)

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

= 1/30

P(case #2) = P(1st ball is red AND 2nd ball is not red AND 3rd ball is yellow)

= P(1st ball is red) x P(2nd ball is not red) x P(3rd ball is yellow)

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

= 1/15

So, P(2nd ball is not red AND 3rd ball is yellow) = P(case #1 OR case #2)

= P(case #1) + P(case #2)

= 1/30 + 1/15

= 1/30 + 2/30

= 3/30

= 1/10

-----------------------------

Here's a different (faster) solution: https://gmatclub.com/forum/a-box-contains-1-blue-ball-1-green-ball-1-yel...

Cheers,

Brent

## Hi,

As per the solution in the Gmatclub :

"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)"

I didnot get what is exactly meant, if we consider 2 ball is yellow than what about the third ball?

and if we are considering 1st ball as not red, meaning it could be green or blue and the third ball is red as 2nd ball is considered yellow.

So can you explain plz

## I'll start with a different

I'll start with a different example:

Let's say a bag contains 3 white balls and 1 red ball, and we start removing balls one at a time (without replacement).

In this case, P(1st ball is red) = P(2nd ball is red) = P(3rd ball is red) = P(4th ball is red)

So, for example, if there are 4 people, and each person removes a ball from the bag, then each person has the SAME likelihood of selecting the red ball. In other words, it doesn't matter WHEN a certain person removes a ball.

The probability that a certain person selects the red ball is always 1/4

Now, let's expand this to the question at hand.

We want to determine P(2nd ball is not red AND 3rd ball is yellow)

Let's first focus on the 2nd ball.

Applying the same logic that we used above, P(2nd ball is not red) = P(1st ball is not red)

Likewise, P(3rd ball is yellow) = P(2nd ball is yellow)

So, P(2nd ball is not red AND 3rd ball is yellow) = P(1st ball is not red AND 2nd ball is yellow)

Does that help?

Cheers,

Brent

## Many thanks for the detailed

## Hi Brent!

https://gre.myprepclub.com/forum/qotd-17-of-the-40-specimens-of-bacteria-in-a-dish-3-speci-2637.html

I was attempting this question and wasn't successful. I would like to know why are we doing 37C4 in the numerator?

Thank you so much for taking some time to answer :)

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/qotd-17-of-the-40-specimens-of-bacteria...

I should mention that the answer choices in the question are not meant to be fractions.

For example, 5/1 is meant to represent 5C1, which can also be represented by a 5 hovering over a 1 (as we see here: https://gre.myprepclub.com/forum/qotd-17-of-the-40-specimens-of-bacteria...)

In order to have exactly 1 specimen with the trait, we must select 1 of the specimens with the trait.

There are 3 specimens with the trait.

So, we can select 1 of these specimens in 3C1 ways.

Now that we have 1 specimen with the trait, the remaining 4 specimens must NOT have the trait.

There are 37 specimens that DON'T have the trait.

So, we can select 4 of these specimens in 37C4 ways.

So, the TOTAL number of ways to select 5 specimens so that 1 has the trait and 4 DON'T have the trait = (3C1)(37C4)

Does that help?

## As easy as it seems, I couldn

## Hi Brent,

I think, this is silly, but I'm confused.

In this question, if the question had asked us to find the number of ways, in which two balls of same color can be randomly selected without replacement, what'd be the answer?

I mean, would you use fundamental counting principle or combination?

Thanks!

Anil

## We have 4 green balls, 3 red

We have 4 green balls, 3 red balls and 2 yellow balls.

So, to get two MATCHING colors, we must consider 3 cases:

i) get 2 green balls

There are 4 green balls. Since the order in which we select the balls does not matter, we can use COMBINATIONS.

We can select 2 green balls from 4 green balls in 4C2 ways (= 6 ways)

ii) get 2 red balls

We can select 2 red balls from 3 red balls in 3C2 ways (= 3 ways)

iii) get 2 yellow balls

We can select 2 yellow balls from 2 yellow balls in 2C2 ways (= 1 way)

So, total number of ways to get same color = 6 + 3 + 1 = 10

Cheers,

Brent

## Thanks for replying, but can

I used fundamental counting principle.

Green: 4x3 = 12 (4 choices for first time and 3 for second)

Red : 3x2 = 6 (3 choices for first time and 2 for second)

Yello: 2x1= 2 (2 choices for first time and 1 for second)

Adding them, total, it comes 20 ways.

I'd be glad if you could explain me, why I couldn't use this counting principle.

## If you use the FCP, then you

If you use the FCP, then you are saying that order matters.

That's fine.

If you assume order matters when calculating the numerator, then you must also do so when calculating the denominator.

So, we can select the first ball in 9 ways

And we can select the second ball in 8 ways

TOTAL = (9)(8) = 72

So, P(same colors) = 20/72 = 5/18

----------------------------------

In my earlier response, we assumed that order didn't matter, and used combinations.

There, we calculated the numerator to be 10

When use combinations to calculate the denominator, we see that we can select 2 balls from 9 balls in 9C2 ways.

9C2 = 36

So, P(same colors) = 10/36 = 5/18

Same answer either way.

Cheers,

Brent

## Great question. If the

## If the question asked for P

If the question asked for P(the balls are different colors), it would take an additional 5 seconds to solve the question.

P(different colors) = 1 - P(same colors)

= 1 - 5/18

= 13/18

More on the complement here: https://www.greenlighttestprep.com/module/gre-probability/video/745