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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 ?

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.

Two integers will 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

I want to understand how will we calculate the total outcomes.

Two integers will 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

I want to understand how will we calculate the total outcomes.

Merging similar topics. Please refer to the solutions above.

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

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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 ?

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

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06 Nov 2015, 19:29

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

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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

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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04 May 2017, 19:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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