# Question: 2nd Ball is Yellow

## Comment on 2nd Ball is Yellow

### How can we solve the 'same

How can we solve the 'same question' with having 2 yellow balls in the group not just one?﻿

### In your REVISED question, we

In your REVISED question, we now have 12 green, 12 red and 2 yellow
We'd have to consider two separate cases:
A) Yellow on 1st pick and yellow on 2nd pick
B) Non-yellow on 1st pick and yellow on 2nd pick

P(A) = (2/26)(1/25)
P(B) = (24/26)(2/25)﻿
So, P(yellow on 2nd pick) = P(A or B)
= P(A) + P(B)
= 2/650 + 48/650
= 50/650
= 1/13﻿

### When it comes to probability

When it comes to probability I find it easier to divide by GCF if any. if two yellow, then divide everything by 2. it becomes 6 G, 6 R, and 1 Y.
=P(A not selecting Yellow)*P(B|A)= 12/13 *1/12=1/13

### Good point, YTarhouni!

Good point, YTarhouni!

ASIDE: I do that as well, nut if I write 24/26, people have a better chance of knowing how I got that number.

### I'm concerned I might get

I'm concerned I might get confused with this kind of problem, because I can't find the logic in it.
To my understanding, getting the yellow ball on the second draw should have a higher probability of success than getting it on the first draw. (On the second draw, the ball has to be taken from 24, not from 25); however, regardless of the way you solve the equation, whether it is 24/25 x 1/24 or 1/25 x 24/24 you still get the same answer... Please help me fix my logic!!! I am caught up in the point that it can't be the same! because, otherwise I could just disregard the fact that it is being drawn in the second attempt and just put down the probability of getting a yellow ball which is 1/25 without solving anything. THANK YOU!

### That's a common misconception

That's a common misconception.
You're right in that, after the 1st ball is selected, there are only 24 balls remaining, so it SEEMS like there's a better probability of drawing the yellow ball on the 2nd draw.
HOWEVER, it's also possible that the yellow ball is selected on the 1st draw, in which case the probability of getting a yellow ball on the 2nd draw is 0.
Those two cases balance out so that the P(yellow ball on 1st draw) = P(yellow ball on 2nd draw)

### This question reminds me of

This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).

So, someone would hold up n pieces of grass (for n people), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that, by selecting last, his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n people had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.

### Thank you so much! I thought

Thank you so much! I thought I had to combine the problem with permutations. The anecdote is a perfect way to clear my confusion. Thanks again!!!!!!!!

### I solved it in this manner,

I solved it in this manner, can you let me know where i'm wrong :-
prob = (24c1 x 1c1)/25c2
(selecting 1 ball out of 24 and 1 yellow ball)/( selecting any two balls out of 25), the answer coming is twice the official one.

### In your numerator, you are

In your numerator, you are treating the counting task as one in which order matters. You have 24 possible outcomes for the FIRST selection and 1 possible outcomes for the SECOND selection. (aside: with this setup, the numerator = 24 x 1 = 24)

However, in the denominator, you are treating the counting task as one in which order DOES NOT matter. You calculated the number of ways to select any 2 balls, in which the order does not matter.

You must be consistent. The counting task in the denominator must also be treated as one in which order DOES matter.

So, there are 25 possible outcomes for the FIRST selection, and there are 24 possible outcomes for the SECOND selection. So, for the denominator, the total number of outcomes = 25 x 24 = 600

So, P(second ball is yellow) = 24/600 = 1/25

### Thanks !! I didn't know that,

Thanks !! I didn't know that, I have been solving questions this way from some time now, whenever prob questions like this came, i used to take out no. of possibilities in this manner for both numertr and denomintr without thinking about the order :(

for example:- A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

I solved it like this (7c1 x 5c1 x 4c1)/16c3 and answer was correct 8/15. Is the 16c3 used here correct bcoz the balls were drawn simultaneously and had they been drawn one after another it would have been 16 x 15 x 14 ?

I think I have a major flaw somewhere in my understanding of the concept.

### Sorry for the delay. This

Sorry for the delay. This question got overlooked somehow.

Your solution (8/15) is correct. However, it is a bit of a fluke.

In your numerator, you are treating the number of outcomes as though order matters. That is, your calculation (7c1 x 5c1 x 4c1) will help us determine the number of ways to select a blue ball then a green ball and then a red ball (IN THAT ORDER).
You must also consider other orders, such as red then blue then green (as well as others). In fact there are 6 different ways to order those three colors. So, your numerator SHOULD BE 6 times bigger (we can arrange 3 objects in 3! ways)

However (fortunately), your denominator SHOULD ALSO be 6 times bigger. If we go with order mattering (as you have done in the numerator), then the denominator should be (16 x 15 x 14). This is the number of ways to select the 1st ball, then the 2nd ball, then the 3rd ball.

So, the solution should be: (7 x 5 x 4 x 3!)/(16 x 15 x 14) = 5/8

Cheers,
Brent

Thanks a lot :)

### Hi Brent,

Hi Brent,

As a general thumbrule, can we say that Numerator and Denominator have to be consistent i.e. if ORDER MATTERS then no matter what the problem statement is, for both Num & Den ORDER SHOULD MATTER. If ORDER does not then its SHOULD NOT MATTER for both Num & Den.

Thanks & Regards,
Abhirup

### Hmmm, that SOUNDS true, and I

Hmmm, that SOUNDS true, and I have a GUT FEELING that it's true. I'm just leery to make it a law :-)
For now, anyway, I can't think of an example where it's not true.

Does that help?

Cheers,
Brent

### Thanks Brent!! That makes

Thanks Brent!! That makes sense.

### An expert on the GMAT Club

An expert on the GMAT Club forums posted this below. Can we use it all the time for situations like this? So without doing the same calculation you did, we recognize that there's one yellow ball out of 25, so answer is 1/25

'Probability of selecting X on the Nth pick is equal to the probability of selecting X on the first pick'

### Yes, that certainly applies

Yes, that certainly applies for this question.

### I am just wandering if its

I am just wandering if its the right way of thinking..25c2=300. then checked all the answer options in which only 25 goes into 300. So chose option c. Is is the right of way thinking to get the answer in quick time?

### That's a great idea,

That's a great idea, santhosh1989. I love the fact that you're looking for a logical shortcut!

The only problem here is that the order of the selected balls matters. For example, selecting red first and yellow second is DIFFERENT FROM selecting yellow first and red second.

So, for this question, the total number of possible outcomes = 600.

The reason for this is that there are 25 ways to select the first ball and then 24 ways to select the second ball, in which case the TOTAL number of outcomes = (25)(24) = 600

So, we can't automatically eliminate A and E

Cheers,
Brent

### Can we also eliminate B

Can we also eliminate B considering that 625>total no of outcomes?

### Great reasoning skills!!

Great reasoning skills!!
Yes, we can definitely eliminate B for that reason.

Cheers,
Brent

### Is the complement of this

Is the complement of this event 'no yellow ball on the second draw'? I am wondering how to solve P(no yellow ball on the second draw)

Thanks a lot! Love your video!

Kate

### Great idea, Kate!

Great idea, Kate!

Let's try it!!

P(2nd draw is yellow) = 1 - P(2nd ball is NOT yellow)

There are two ways for the second ball to be a color other than yellow. EITHER the 1st ball can is yellow (which means the 2nd ball definitely won't be yellow) OR the 1st and 2nd ball are both not yellow.
So, we get:

P(2nd ball is NOT yellow) = P(1st ball is yellow OR 1st ball is not yellow and 2nd ball is not yellow)

= P(1st ball is yellow) + P(1st ball is not yellow and 2nd ball is not yellow)

= P(1st ball is yellow) + [P(1st ball is not yellow) x P(2nd
ball is not yellow)]

= 1/25 + [24/25 x 23/24]

= 1/25 + 23/25

= 24/25

We get: P(2nd draw is yellow) = 1 - 24/25
= 1/25

### Hi Brent I did not understand

Hi Brent I did not understand why didn't you counting principle in the denominator? I was trying to solve the question doing that but failing to get an answer.

### As with many probability

As with many probability questions, we can also solve the question using counting techniques.

P(2nd ball is yellow) = (# of outcomes in which the 2nd ball is yellow)/(TOTAL number of outcomes)

DENOMINATOR: TOTAL number of outcomes
There are 25 balls
We can select the 1st ball in 25 ways
We can select the 2nd ball in 24 ways
So, TOTAL number of outcomes = (25)(24)

NUMERATOR: # of outcomes in which the 2nd ball is yellow
We can select the 1st ball in 24 ways (any ball OTHER than the yellow one)
We can select the 2nd ball in 1 way (this ball MUST be yellow)
Number of outcomes = (24)(1) = 24

So, P(2nd ball is yellow) = 24/(25)(24) = 1/25

Cheers,
Brent