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## Comment on

Sum of 11## Great

## sir i have a doubt here we

## Let's use your analogy with

Let's use your analogy with the dice to illustrate the problem. We have 2 dice, so let's make one die red and one die blue.

With this set-up, we can see that there are two ways to get a sum of 11:

case a: the red die is 5 and the blue die is 6

case b: the red die is 6 and the blue die is 5

IMPORTANT: The two cases above are DIFFERENT.

This set-up, however, does not apply to the question in the above video. Let's call the sets A and B.

A: {1, 2, 4, 6, 7}

B: {2, 3, 4, 5, 6, 7, 8, 9,}

Let's LIST the ways can we get a sum of 11:

case a: select 2 from set A, and select 9 from set B

case b: select 4 from set A, and select 7 from set B

case c: select 6 from set A, and select 5 from set B

case d: select 7 from set A, and select 4 from set B

These are the only ways to get a sum of 11.

Notice that case b and case d are different cases. In one case, we select the 4 from set A and in the other case, we select the 4 from set B.

In your question, you are suggesting that 2+9 and 9+2 are DIFFERENT.

However, there's only ONE way to get a sum with a 9 and a 2: We MUST select a 2 from set A and select a 9 from set B, since it's impossible to select a 9 from set A and select a 2 from set B.

## Hi,

I wanted to ask why are the two cases different, meaning why can't we have a possibility of selecting 9 from set B and 2 from set A?

Is it because first we have to select a number from set A and then one from set B and keep the order?

Thanks

## We could have arranged our

We could have arranged our solution by selecting a number from set B before selecting a number from set A, but the answer would be the same.

The key is to recognize that approaches yield the same outcomes.

For example, notice that the outcome created by selecting 9 from set B and selecting 2 from set A is the same outcome by selecting 2 from set A and selecting 9 from set B.

Does that help?

Cheers,

Brent

## Three positive integers are

a) 1/28

b) 9/29

c) 9/28

d) 1/3

e) 5/14

## Can a number be selected more

There's a nice solution here: http://www.urch.com/forums/gmat-problem-solving/37062-probability-averag...

## I have missed the source but

## Hello Brent,

How is the pair {4,7} different from {7,4}?

Should we not consider as redundant?

## Selecting a 4 from set A and

Selecting a 4 from set A and a 7 from set B is different from a 7 from set A and a 4 from set B.

## Was very clear, thank you. I

But I have another question. If we have N1 & N2 in (N1, N2), would (N1=4, N2=4) be the same as (N2=4, N1=4)? I mean is each pick different, even if you pick the same number from the two?

Thanks.

## Yes, selecting 4 from set A

Yes, selecting 4 from set A and selecting 7 from set B is DIFFERENT FROM selecting 7 from set A and selecting 4 from set B.

However, selecting 4 from set A and selecting 4 from set B is THE SAME AS selecting 4 from set B and selecting 4 from set A.

Cheers,

Brent

## Hi Brent, is there anyway we

## We are choosing only ONE

We are choosing only ONE number from each set, the order definitely doesn't matter.

We can select one number from the first set in 5C1 ways (= 5 ways). Or we can just recognize there are 5 ways to select one number from the first set.

Likewise, we can select one number from the second set in 8C1 ways (= 8 ways). Or we can just recognize there are 8 ways to select one number from the second set.

So, the number of ways to select one number from EACH set = (5)(8) = 40

Does that help?

Cheers,

Brent

## I found the question a bit

## It's important to note that

It's important to note that the order in which we choose two numbers makes no difference to the solution. Likewise, choosing the two numbers simultaneously will also yield the same solution.

## Can i solve this question by

Number of possible pairs of sum 11 are (2 9)or (4 7) or (6 5) or (7 4)

4*(1/5)*(1/8) = 1/10

## Interesting solution! Nice

Interesting solution! Nice work!

Essentially, your approach is as follows:

P(sum is 11) = P(2 then 9 OR 4 then 7 OR 6 then 5 OR 7 then 4)

= P(2 then 9) + P(4 then 7) + P(6 then 5) + P(7 then 4)

= (1/5)(1/8) + (1/5)(1/8) + (1/5)(1/8) + (1/5)(1/8)

= 4(1/5)(1/8)

## Hey, your solution is very

First I choose one number from the first set, there are 4 numbers that can create 11, so the probability is 4/5.

After I chose one from the first set, I have only one option in the second set to get the 11.

So first I choose from set one and then from set two.

(4/5)*(1/8)=1/10

It looks correct, but here is the question.

Shouldn't I multiply it by 2?

There are two possible scenarios, I choose my first ball from the first set and the second scenario is that I choose my first ball from the second set.

I have a feeling that there is no need to multiply it by 2 but I can't explain why. Can you please help me to understand it?

## Your solution is perfectly

Your solution is perfectly valid.

Here's how it breaks down mathematically:

P(sum is 11) = P(1st # is 1 of the 4 #'s that can create a sum of 11 AND 2nd # is the matching # to make 11)

= P(1st # is 1 of the 4 #'s that can create a sum of 11) x P(2nd # is the matching # to make 11)

= 4/5 x 1/8 = 1/10

As you can see, there's no duplication here.

## Thanks!