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## Comment on

At Least 1 Even Number## In this question, the total

## I don't say that the total

I don't say that the total number of outcomes is 10C2. I say it's 5C2.

We have two options here. We can say that order does not matter in both the numerator and denominator (as I did in my solution). Or we can say that order DOES matter in both the numerator and denominator (as you are suggesting). As you say, however, the answer is the same in both instances.

## can we answer it like:

Therefore the ans =1-1/20=18/20.

is it possible?

## Close! First, 1 - 1/20 does

Close! First, 1 - 1/20 does not equal 18/20. So, we have a problem there :-)

Your calculation of (1/5)(1/4) = 1/20 represents the probability of selecting the 3 FIRST and the 5 SECOND.

We must also consider the probability of selecting the 5 FIRST and the 3 SECOND. This too is equal to (1/5)(1/4) = 1/20

So, P(selecting 3 and 5) = 1/20 + 1/20 = 1/10

So, P(selecting at least one even) = 1 - 1/10 = 9/10

## Can we solve this problem

We have to select two numbers (Num1 and Num2). There is "and" sign it , It means multiplication. if we select two odd numbers (2/5*1/4)=1/10. Then P(even No:) = 1-1/10=0.9.

Need you Acknowledgment for this procedure.

## Yes, that's a very valid

Yes, that's a very valid solution. This practice question appears early in the probability module and is meant to reinforce some of the more basic probability strategies.

## Hi,

In your previous video about "p( product of 3 numbers will be odd)", you found out the total no: of outcomes in the denominator by choosing 3 no: from each of the 3 sets, which is 2x5x4, which could be interpreted as 2C1x5C1x4C1.

But why can't I use the same interpretation when it comes to choosing 2 no:s from a SINGLE set to get the denominator as 5C1x4C1 instead of 5C2? although I realize in this case order of the no's are inadvertently taken into consideration, can you give me a sound reasoning as to why that logic/method is forbidden? why am I not allowed to write here 5C1x4C1 while in the former I was able to split into 2C1x5C1x4C1? Is it because in the former it was 3 distinct sets and not allowed for a single set?

## Your question is similar to

Your question is similar to the question that Rohan asked at the top of this thread (so my answer will be similar).

We have two options here. We can say that order does not matter in both the numerator and denominator (as I did in my solution). Or we can say that order DOES matter in both the numerator and denominator (as you are suggesting). As you say, however, the answer is the same in both instances.

If you want to say that order matters when calculating the denominator, the we must say that order matters when calculating the numerator.

You have already calculated the denominator (with the premise that order matter), and you got 5C1 x 4C1 = 5 x 4 = 20

Now the denominator. In how many ways can we select 2 of the numbers so that we don't get any even numbers?

There are only 2 odd (i.e., non-even numbers).

So, there are 2 ways (2C1 if you wish) to select the first odd number, and there is 1 way to select the second odd number). So, in TOTAL, the number of ways to select 2 odd numbers = 2 x 1 = 2

So, P(select zero even numbers) = 2/20 = 1/10

So, P(at least 1 even number) =- 1 - 1/10 = 9/10

## Why can't you do 2/5C2? Why

## Please see my response to

Please see my response to ananthu's question above.

## Hi Brent,by following your

## Hi Runnerboy44,

Hi Runnerboy44,

Yes, we can select 2 even numbers in 3C2 ways. However, I'm not sure what you want to do with this information.

The question asks us to find P(at least one even number)

There are two possible cases that give us at least one even number:

CASE 1: one number is even and the other number is odd

CASE 2: both numbers are even

So, 3C2 = the number of ways to accomplish CASE 2, but you still need to address CASE 1.

Does that help?

Cheers,

Brent

## yes it does help Brent. When

## Perfect!

Perfect!

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