Frequency Distribution of Integers in Set X and Set YX and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?a)1/15b)13/354c)1/5d)1/2e)4/5Correct answer is option 'D'. Can you explain this answer?

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Frequency Distribution of Integers in Set X and Set Y

X and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?

a)
1/15
b)
13/354
c)
1/5
d)
1/2
e)
4/5

Correct answer is option 'D'. Can you explain this answer?

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Frequency Distribution of Integers in Set X and Set YX and Y are two s...

Given:

Sets X and Y contain integers as shown in the table.
1 integer is chosen from X (the integer chosen from Set X will be referred to as I_X from now on) and one from Y (I_Y)

To find: The probability that I_X*I_Y is Even

Approach:

P(I_X*I_Y is Even) = 1 – P(I_X*I_Y is Odd)
- The reason why we are taking the Non-Event Approach here is that there are 3 ways in which the product of 2 integers can be Even ( i. Both integers are even ii. Only I_X is even and iii. Only I_Y is even). However, there is only one way in which I_X*I_Y is odd (when both the integers are odd). So, it’s easier and quicker to solve the question using the Non-Event Approach.
Since the product I_X*I_Y is odd when I_X and I_Y are odd, we can write:
- P(I_X*I_Y is Odd) = P(I_X is odd)*P(I_Y is odd)
So, we need to find P(I_X is odd) and P(I_Y is odd)

Working Out:

Finding P(I_X is odd)

The total number of integers in Set X is 30

So, the total number of ways in which one integer can be selected from Set X =
The odd integers in Set X (3, 5, 7) are highlighted in the table. 3 occurs 4 times in Set X, 5 occurs 6 times and 7 occurs 8 times
So, the total number of odd integers in Set X = 4 + 6 + 8 = 18
So, the number of ways in which one ODD integer can be selected from Set X=
Therefore, P(I_X is odd) =
Finding P(I_Y is odd)
The total number of integers in Set Y is 12
So, the total number of ways in which one integer can be selected from Set
All the integers in Set Y are odd (highlighted in the table) except the two occurrences of 24.
So, the total number of odd integers in Set Y = 12 – 2 = 10
So, the number of ways in which one ODD integer can be selected from Set Y =
Therefore, P(I_Y is odd) =

Finding the Required Probability

Looking at the answer choices, we see that the correct answer is Option D

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Most Upvoted Answer

Frequency Distribution of Integers in Set X and Set YX and Y are two s...

Given:

Sets X and Y contain integers as shown in the table.
1 integer is chosen from X (the integer chosen from Set X will be referred to as I_X from now on) and one from Y (I_Y)

To find: The probability that I_X*I_Y is Even

Approach:

P(I_X*I_Y is Even) = 1 – P(I_X*I_Y is Odd)
- The reason why we are taking the Non-Event Approach here is that there are 3 ways in which the product of 2 integers can be Even ( i. Both integers are even ii. Only I_X is even and iii. Only I_Y is even). However, there is only one way in which I_X*I_Y is odd (when both the integers are odd). So, it’s easier and quicker to solve the question using the Non-Event Approach.
Since the product I_X*I_Y is odd when I_X and I_Y are odd, we can write:
- P(I_X*I_Y is Odd) = P(I_X is odd)*P(I_Y is odd)
So, we need to find P(I_X is odd) and P(I_Y is odd)

Working Out:

Finding P(I_X is odd)

The total number of integers in Set X is 30

So, the total number of ways in which one integer can be selected from Set X =
The odd integers in Set X (3, 5, 7) are highlighted in the table. 3 occurs 4 times in Set X, 5 occurs 6 times and 7 occurs 8 times
So, the total number of odd integers in Set X = 4 + 6 + 8 = 18
So, the number of ways in which one ODD integer can be selected from Set X=
Therefore, P(I_X is odd) =
Finding P(I_Y is odd)
The total number of integers in Set Y is 12
So, the total number of ways in which one integer can be selected from Set
All the integers in Set Y are odd (highlighted in the table) except the two occurrences of 24.
So, the total number of odd integers in Set Y = 12 – 2 = 10
So, the number of ways in which one ODD integer can be selected from Set Y =
Therefore, P(I_Y is odd) =

Finding the Required Probability

Looking at the answer choices, we see that the correct answer is Option D

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Community Answer

Frequency Distribution of Integers in Set X and Set YX and Y are two s...

Given:

Sets X and Y contain integers as shown in the table.
1 integer is chosen from X (the integer chosen from Set X will be referred to as I_X from now on) and one from Y (I_Y)

To find: The probability that I_X*I_Y is Even

Approach:

P(I_X*I_Y is Even) = 1 – P(I_X*I_Y is Odd)
- The reason why we are taking the Non-Event Approach here is that there are 3 ways in which the product of 2 integers can be Even ( i. Both integers are even ii. Only I_X is even and iii. Only I_Y is even). However, there is only one way in which I_X*I_Y is odd (when both the integers are odd). So, it’s easier and quicker to solve the question using the Non-Event Approach.
Since the product I_X*I_Y is odd when I_X and I_Y are odd, we can write:
- P(I_X*I_Y is Odd) = P(I_X is odd)*P(I_Y is odd)
So, we need to find P(I_X is odd) and P(I_Y is odd)

Working Out:

Finding P(I_X is odd)

The total number of integers in Set X is 30

So, the total number of ways in which one integer can be selected from Set X =
The odd integers in Set X (3, 5, 7) are highlighted in the table. 3 occurs 4 times in Set X, 5 occurs 6 times and 7 occurs 8 times
So, the total number of odd integers in Set X = 4 + 6 + 8 = 18
So, the number of ways in which one ODD integer can be selected from Set X=
Therefore, P(I_X is odd) =
Finding P(I_Y is odd)
The total number of integers in Set Y is 12
So, the total number of ways in which one integer can be selected from Set
All the integers in Set Y are odd (highlighted in the table) except the two occurrences of 24.
So, the total number of odd integers in Set Y = 12 – 2 = 10
So, the number of ways in which one ODD integer can be selected from Set Y =
Therefore, P(I_Y is odd) =

Finding the Required Probability

Looking at the answer choices, we see that the correct answer is Option D

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Frequency Distribution of Integers in Set X and Set YX and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?a)1/15b)13/354c)1/5d)1/2e)4/5Correct answer is option 'D'. Can you explain this answer?

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Frequency Distribution of Integers in Set X and Set YX and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?a)1/15b)13/354c)1/5d)1/2e)4/5Correct answer is option 'D'. Can you explain this answer? for GMAT 2024 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about Frequency Distribution of Integers in Set X and Set YX and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?a)1/15b)13/354c)1/5d)1/2e)4/5Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GMAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Frequency Distribution of Integers in Set X and Set YX and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?a)1/15b)13/354c)1/5d)1/2e)4/5Correct answer is option 'D'. Can you explain this answer?.

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