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

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

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

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.

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

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

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.



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.

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.

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.

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.

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

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.

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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.

Here's a quick solution.

Key property: Notice that, for every number in set A, there is ONE number in set B that will produce a sum of 9.

So, if you randomly select a number from set A ([I suggest that you actually do this at home]), you can see that, among the 5 numbers in set B, only 1 of them will combine with your number from set A to create a sum of 9.

So, the probability of getting a sum of 9 = 1/5 = 0.2

Answer: B

Cheers,
Brent

Solution

Given

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

To Find

• Probability of selecting two integers, one from each set A and B such that the sum = 9.

Approach and Working out
\(Probability = \frac{Favourable-cases}{ Total-cases}.\)

Favourable cases:

• To get the sum = 9, we have to select numbers in such a way that together the elements from A and B make up the sum = 9.
• Possible pairs will be (2, 7), (3, 6), (4, 5) only.

o For the pair (2, 7): only one possibility is there that is 2 should come from A and 7 from B. Therefore number of favourable cases = 1.

 Since 2 doesn’t exist in set B.
o For the pair (3, 6): only one possibility is there that is 3 should come from A and 6 from B. Therefore number of favourable cases = 1
o For the pair (4, 5): there are two possibilities.

 If 4 comes from A, 5 comes from B. If 4 comes from B, 5 comes from A.
• Total number of favourable cases = 2.

Total number of favourable cases = 1 + 1 + 2 = 4.
Total number of cases = selecting one from each A and B = \(4C_1*5C_1\) = 20
Probability =\( \frac{4}{20}\) = 0.20.

Correct Answer: Option B