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# Two integers will be randomly selected from the sets above, one intege

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

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Updated on: 31 Jan 2019, 00:55
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Question Stats:

79% (01:29) correct 21% (01:46) wrong based on 1336 sessions

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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

Originally posted by vermatanya on 17 Jan 2008, 13:50.
Last edited by Bunuel on 31 Jan 2019, 00:55, edited 2 times in total.
Edited the question.
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Posts: 65771
Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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24 Feb 2012, 21:49
12
5
fortsill wrote:
walker wrote:
B

$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$

or

$$p=\frac{4}{4}*\frac{1}{5}=0.20$$

or

$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$

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.

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.

Hope it's clear.
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Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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07 Sep 2009, 10:29
8
3
Maybe I am late with response, but anyway.

We have two sets with 4 and 5 integers included in each one respectively.

Thus, we have 4*5=20 possible ways of selecting two integers from these sets.

The requirement is that we need the sum of 9.
Here we have 4 possible choices:

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

Probability = favourable / total

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

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17 Jan 2008, 13:59
3
4
B

$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$

or

$$p=\frac{4}{4}*\frac{1}{5}=0.20$$

or

$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$
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Posts: 418
Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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17 Jan 2008, 14:26
3
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.
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Joined: 04 Jan 2008
Posts: 47
Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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22 Jan 2008, 16:10
2
B
total 20 ways to sum,
4 of the sums equals 9
so, prob is 4/20= 1/5= .20
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Joined: 24 Feb 2012
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Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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24 Feb 2012, 20:54
walker wrote:
B

$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$

or

$$p=\frac{4}{4}*\frac{1}{5}=0.20$$

or

$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$

Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?
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Joined: 08 Jun 2011
Posts: 72
Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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03 May 2012, 04:11
Bunuel wrote:
fortsill wrote:
walker wrote:
B

$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$

or

$$p=\frac{4}{4}*\frac{1}{5}=0.20$$

or

$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$

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.

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.

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)?

if-two-of-the-four-expressions-x-y-x-5y-x-y-5x-y-are-92727.html#p713823
Thank you
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Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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03 May 2012, 08:29
1
2

2 from 9 is 36
and
2 from 7 is 21

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.

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

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20 Aug 2012, 03:12
Responding to a pm:

No, there will not be 8 different pairs.

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

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

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15 Nov 2012, 02:25
HI Karishma,

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

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15 Nov 2012, 02:33
HI Karishma,

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

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15 Nov 2012, 02:40
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??
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Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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15 Nov 2012, 02:46
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.
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Re: Two integers will be randomly selected from the sets above, one intege  [#permalink]

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02 Dec 2012, 17:51
7

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, one intege  [#permalink]

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06 Jul 2013, 07:22
walker wrote:
B

$$p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20$$

or

$$p=\frac{4}{4}*\frac{1}{5}=0.20$$

or

$$p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20$$

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

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03 May 2016, 04:46
2
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, one intege  [#permalink]

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22 Feb 2018, 13:04
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, one intege  [#permalink]

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10 Apr 2018, 05:45
Bunuel 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.

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, one intege  [#permalink]

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10 Apr 2018, 05:51
@s wrote:
Bunuel 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.

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.
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Re: Two integers will be randomly selected from the sets above, one intege   [#permalink] 10 Apr 2018, 05:51

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