Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Two integers will be randomly selected from the sets above [#permalink]

Show Tags

17 Jan 2008, 14:50

3

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

82% (00:43) correct 18% (00:53) wrong based on 552 sessions

HideShow timer Statistics

A = {2, 3, 4, 5} B = {4, 5, 6, 7, 8}

Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ?

You don't need to perform any calculations in this example. Regardless of the number you choose from set A, there is only one number (out of 5) that makes the sum equal 9. Therefore, the probability is 1/5. The answer is B.

2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?

A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33

-thanx

p= 4/(4c1*5c1)=1/5=0.2
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?

A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33

-thanx

Total ways of choosing 1 number from Set A and 1 from Set B.... 4c1 x 5c1 = 20 Comb which yield 9 as the sum - 2,7.....3,6.....4,5....5,4 = 4 comb...

Probability = 4 / 20 = .20 = B
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?

Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?

I'd offer another approach which I hope will also clarify the above.

A = {2,3,4,5} B = {4,5,6,7,8}

2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33

Rearrange the first set: A = {5, 4, 3, 2} B = {4, 5, 6, 7, 8}

As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2.

Answer: B.

Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5.

Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?

I'd offer another approach which I hope will also clarify the above.

A = {2,3,4,5} B = {4,5,6,7,8}

2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33

Rearrange the first set: A = {5, 4, 3, 2} B = {4, 5, 6, 7, 8}

As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2.

Answer: B.

Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5.

Answer: B.

Hope it's clear.

Bunuel,

I am confused.

We are selecting a pair from a total of 9 numbers or 7 numbers (if we eliminate duplicates). So we will be getting for total number of ways, in either case

2 from 9 is 36 and 2 from 7 is 21

So we have [ (2, 7),(3,6), (4,5) ] 3 pairs or

[ (2, 7),(3,6), (4,5) (5,4) ] 4 pairs if we don't eliminate duplicates

Either way, I am getting the wrong answer

3/36 and 4/21 are both wrong.

Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same.

How is it different from this one (the general concept)?

You are not selecting 2 numbers from 9 numbers. You are selecting 1 number from 4 numbers and 1 number from 5 numbers. Mind you, they are different.

While selecting 2 numbers from {1, 2, 3, 4, 5, 6, 7, 8, 9}, you can select (3, 4) While selecting 1 from 4 and 1 from 5: {1, 2, 3, 4} and {5, 6, 7, 8, 9}, you cannot select (3, 4).

That's the problem number 1.

Lstadt wrote:

Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same.

Secondly, the two given sets are distinct: A = {2,3,4,5} B = {4,5,6,7,8}

When we pick a number from A and from B, we get {4, 5} or we can do it in this way {5, 4}. These are two different cases. In the first case, the 4 comes from A and in the second case, it comes from B.
_________________

If you select 2 from A, from B you must select 7. If you select 3 from A, from B you must select 6. If you select 4 from A, from B you must select 5. If you select 5 from A, from B you must select 4. The total number of ways in which the sum can be 9 is 4.

The question was whether the last two cases are distinct or not. Is 4 from A and 5 from B the same as 4 from B and 5 from A. Since 4 and 5 come from different sets in the two cases, they represent two different cases and should be counted as such.
_________________

I understood the question, got 0.20, but I have a doubt.

At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.

How do we come too know that we need not include duplicate pairs??

I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q?

You need to select a pair {a, b} such that their sum is 9. a is selected from set A and b is selected from set B.

a can take a value from set A only i.e. one of 2/3/4/5 b can take a value from set B only i.e. one of 4/5/6/7/8

How do you select {6, 3}? a cannot be 6. A selection is different from another when you select different numbers from each set. One selection is {4, 5} - 4 from A, 5 from B Another selection is {5, 4} - 5 from A, 4 from B

One selection is {3, 6} - 3 from A, 6 from B There is no such selection {6, 3} - A has no 6.
_________________

Re: 2 integers will be randomly selected from the sets above [#permalink]

Show Tags

15 Nov 2012, 03:40

when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order??

when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order??

It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter. {4, 5} - 4 is from A and 5 is from B {5, 4} - 5 is from A and 4 is from B These two are different.

{2, 6} - 2 is from A and 6 is from B {6, 2} - 2 is from A and 6 is from B These two are same.
_________________

I made it right, but I don´t understand this approach (Combinatorics). Why the denominator in the first solution \(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\) is \(\ C^4_1\) Some help? Thanks!
_________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

I made it right, but I don´t understand this approach (Combinatorics). Why the denominator in the first solution \(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\) is \(\ C^4_1\) Some help? Thanks!

A = {2, 3, 4, 5} B = {4, 5, 6, 7, 8}

Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ?

(A) 0.15 (B) 0.20 (C) 0.25 (D) 0.30 (E) 0.33

The probability = (# favorable outcomes)/(total # of outcomes)

(# favorable outcomes) is 4C1=4: (2, 7); (3, 6), (4, 5) and (5, 4). (total # of outcomes) is 4C1*5C1=4*5=20.