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Two integers will be randomly selected from the sets above,

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

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

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Walkabout wrote:
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 total number of pairs possible is 4*5=20. Out of these 20 pairs only 4 sum up to 9: (2, 7); (3, 6), (4, 5) and (5, 4). The probability thus is 4/20=0.2.

Answer: B.
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Re: Two integers will be randomly selected from the sets above, [#permalink]

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New post 02 Dec 2012, 18:51
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How about this line of thought:

I pick a number from Set A. No matter which number I pick, my chance of chosing the right number (which gives A+B = 9) in Set B will be 1/5.

Therefore 1/5 = 0.20 => B
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Re: Two integers will be randomly selected from the sets above, [#permalink]

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New post 11 Sep 2014, 01:54
Walkabout wrote:
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



total possibility -- 4c1 *5c1

possible stuff: (2,7), (3,6), (4,5),(5,4)= 4

4/20= 1/5= 0.2
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Re: Two integers will be randomly selected from the sets above, [#permalink]

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New post 02 May 2016, 15:23
Probability of A (any number) AND B(4 of 5).
We don't need to calculate each and every probability, rather the overall. But let's go the long way to see the theory behind the easy way.

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

If we choose 2 from A, the only way we can get a sum of 9 is if we choose 7 from B.

2 from A = .25 probability of occurring (1/4)
7 from B = .20 probability of occurring (1/5)
(1/4)*(1/5) = 1/20
If you do this for each number in A, you get the same outcome, 1/20
add all of these up, and you get 1/20 + 1/20 + 1/20 + 1/20 = 4/20 = 4/20(5) = 20/100 = .2 to get the sum of 9 when choosing a number from A and its corresponding "match" from B (3+5)(2+7)(4+5)(5+4)

Now to take the easy route.
For any number in A, you have a 4/4 chance of picking a number (100% strike rate)
For any number in B, you have a 4/5 chance of picking a number (80% strike rate)
4/4 * 4/5 = 16/20 chance of getting any addition answer, therefore a 4/20 chance of getting 9 as a sum = 20%, .2
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Re: Two integers will be randomly selected from the sets above, [#permalink]

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New post 03 May 2016, 05:46
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Walkabout wrote:
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


Solution:

To determine the probability that the sum of the two integers will equal 9, we must first recognize that probability = (favorable outcomes)/(total outcomes).

Let’s first determine the total number of outcomes. We have 4 numbers in set A, and 5 in set B, and since we are selecting 1 number from each set, the total number of outcomes is 4 x 5 = 20.

For our favorable outcomes, we need to determine the number of ways we can get a number from set A and a number from set B to sum to 9. We are selecting from the following two sets:

A = {2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

We will denote the first number as from set A and the second from set B. Here are the pairings that yield a sum of 9:

2,7
3,6
4,5
5,4

We see that there are 4 favorable outcomes. Thus, our probability is 4/20 = 0.25, Answer C.
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Re: Two integers will be randomly selected from the sets above, [#permalink]

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New post 22 Feb 2018, 14:04
Hi All,

Probability questions are based on the probability formula:

(Number of ways that you "want") / (Total number of ways possible)

Since it's usually easier to calculate the total number of possibilities, I'll do that first. There are 4 options for set A and 5 options for set B; since we're choosing one option from each, the total possibilities = 4 x 5 = 20

Now, to figure out the number of duos that sum to 9:

2 and 7
3 and 6
4 and 5
5 and 4

4 options that give us what we "want"

4/20 = 1/5 = 20% = .2

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

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New post 10 Apr 2018, 06:45
Bunuel wrote:
Walkabout wrote:
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 total number of pairs possible is 4*5=20. Out of these 20 pairs only 4 sum up to 9: (2, 7); (3, 6), (4, 5) and (5, 4). The probability thus is 4/20=0.2.

Answer: B.



Hi Bunuel

Is {4,5} and {5,4} not same?
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Re: Two integers will be randomly selected from the sets above, [#permalink]

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New post 10 Apr 2018, 06:51
@s wrote:
Bunuel wrote:
Walkabout wrote:
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 total number of pairs possible is 4*5=20. Out of these 20 pairs only 4 sum up to 9: (2, 7); (3, 6), (4, 5) and (5, 4). The probability thus is 4/20=0.2.

Answer: B.



Hi Bunuel

Is {4,5} and {5,4} not same?



(4, 5) is 4 from A and 5 from B.

(5, 4) is 5 from A and 4 from B.

Those are two different cases.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: Two integers will be randomly selected from the sets above,   [#permalink] 10 Apr 2018, 06:51
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