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655-705 Level|   Probability|      
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daviesj
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Set A consists of all the integers between 10 and 21, inclus 28 Dec 2012, 07:18
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B
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Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4
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B
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Set A consists of all the integers between 10 and 21, inclus 28 Dec 2012, 20:34
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From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4

Therefore probability = 4/12 = 1/3

Option B
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Set A consists of all the integers between 10 and 21, inclus 03 Jan 2013, 05:28
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If y has no factor z such that 1 < z < y, then y must be prime. Let's look at a few examples to
see why this is true:
6 has a factor 2 such that 1 < 2 < 6: 6 is NOT prime
15 has a factor 5 such that 1 < 5 < 15: 15 is NOT prime
3 has NO factor between 1 and 3: 3 IS prime
7 has NO factor between 1 and 7: 7 IS prime
Because it is selected from Set B, y is a prime number between 10 and 50, inclusive. The only prime
number that is divisible by 3 is 3, so y is definitely not divisible by 3.
Thus, xy is only divisible by 3 if x itself is divisible by 3. We can rephrase the question: “What is
the probability that a multiple of 3 will be chosen randomly from Set A?”
There are 21 – 10 + 1 = 12 terms in Set A. Of these, 4 terms (12, 15, 18, and 21) are divisible by 3

Thus, the probability that x is divisible by 3 is 4/12 = 1/3.
The correct answer is B. (Source : Manhattan GMAT Advanced Quant Guide)
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Set A consists of all the integers between 10 and 21, inclus 21 Nov 2023, 07:30
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GyanOne
From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4

Therefore probability = 4/12 = 1/3

Option B


Can you please explain why you choose 12 as the base?
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Set A consists of all the integers between 10 and 21, inclus 21 Nov 2023, 08:07
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pooja.sharma2909
GyanOne
Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4

Therefore probability = 4/12 = 1/3

Option B


Can you please explain why you choose 12 as the base?

Because there are 12 integers between 10 and 21, inclusive.

Essentially, the question asks: If we choose an integer from 10 to 21, inclusive, and another integer from the prime numbers between 10 and 50, what is the probability that their product will be a multiple of 3?

Since among the prime numbers from 10 to 50, there are no multiples of 3 (the only prime which is a multiple of 3 is 3), then for the product to be a multiple of 3, the number chosen from 10 to 21 must be a multiple of 3. There are a total of 12 integers in that range, out of which 4 are multiples of 3: 12, 15, 18, and 21. Therefore, the probability is 4/12 = 1/3.

Answer: B.

Hope it helps.
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Set A consists of all the integers between 10 and 21, inclus 22 Nov 2023, 23:19
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Number of elements in Set A = 12
Number of elements in Set B = 41

x can be any of the 12 values in set A
y can be any of the 11 primes in Set B (11,13,17,19,23,29,31,37,41, 43 and 47)

Total number of outcomes = 12 x 11 =132

xy must be divisible by 3 -
Total number of favorable outcomes will be with x=12, 15, 18 or 21, that is 4 values of x and 11 values of y
Therefore tota number of favorabe outcomes = 4 x 11 = 44

Required probability = 44/132 = 1/3
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Set A consists of all the integers between 10 and 21, inclus 25 Apr 2024, 04:02
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Bunuel
pooja.sharma2909
GyanOne
Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4

Therefore probability = 4/12 = 1/3

Option B

Can you please explain why you choose 12 as the base?
Because there are 12 integers between 10 and 21, inclusive.

Essentially, the question asks: If we choose an integer from 10 to 21, inclusive, and another integer from the prime numbers between 10 and 50, what is the probability that their product will be a multiple of 3?

Since among the prime numbers from 10 to 50, there are no multiples of 3 (the only prime which is a multiple of 3 is 3), then for the product to be a multiple of 3, the number chosen from 10 to 21 must be a multiple of 3. There are a total of 12 integers in that range, out of which 4 are multiples of 3: 12, 15, 18, and 21. Therefore, the probability is 4/12 = 1/3.

Answer: B.

Hope it helps.
Bunuel
Questions says ''y is a number chosen randomly from Set B''

So, why haven't we considered accounting the probablity of y = (11/41)? why is that probab. of choosing ''y'' will be = 1?
Please help!!
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Set A consists of all the integers between 10 and 21, inclus 25 Apr 2024, 06:40
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Contropositive
Bunuel
Questions says ''y is a number chosen randomly from Set B''

So, why haven't we considered accounting the probablity of y = (11/41)? why is that probab. of choosing ''y'' will be = 1?
Please help!!
y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y means that y is a prime number. So, y is already one one of the numbers from {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}. Thus, we need (a multiple of 3 from A)*(any number from possible values of y) =­ 4/12*1.­
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Set A consists of all the integers between 10 and 21, inclus 25 Apr 2024, 07:03
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Should the denominator be 12 + 41? Number x is chosen from set A, so you have 12 options. Number y is chosen from set B, so you have 41 options.

Bunuel No?

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Set A consists of all the integers between 10 and 21, inclus 25 Apr 2024, 07:28
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hughng92
Should the denominator be 12 + 41? Number x is chosen from set A, so you have 12 options. Number y is chosen from set B, so you have 41 options.

Bunuel No?

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­I understand the confusion. However, this is what the question means:

x is chosen from {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}
y is chosen from {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}

What is the probability that xy is a multiple of 3?

For the product to be a multiple of 3, x must be a multiple of 3 because there are no multiples of 3 in the second set. Hence, the probability is 4/12 = 1/3. 
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Set A consists of all the integers between 10 and 21, inclus 25 Apr 2024, 07:31
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I see. If I phrase the second part of selection such as “Pick y from the set B and also y does not contain a factor z such that 1 < z < y”, then I will arrive at the correct number of denominator.

Makes sense. Thanks!

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Set A consists of all the integers between 10 and 21, inclus 12 May 2024, 09:49
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Bunuel
pooja.sharma2909
GyanOne
Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4

Therefore probability = 4/12 = 1/3

Option B

Can you please explain why you choose 12 as the base?
Because there are 12 integers between 10 and 21, inclusive.

Essentially, the question asks: If we choose an integer from 10 to 21, inclusive, and another integer from the prime numbers between 10 and 50, what is the probability that their product will be a multiple of 3?

Since among the prime numbers from 10 to 50, there are no multiples of 3 (the only prime which is a multiple of 3 is 3), then for the product to be a multiple of 3, the number chosen from 10 to 21 must be a multiple of 3. There are a total of 12 integers in that range, out of which 4 are multiples of 3: 12, 15, 18, and 21. Therefore, the probability is 4/12 = 1/3.

Answer: B.

Hope it helps.
­Question:
In the question, y is also randomly selected which should meet the condition of y to be a prime number from set B. Therefore, along with the selection of x (i.e., multiple of 3 from set A), shouldn't there be selection of y as prime numbers from set B?

Probability of selecting multiple of 3 from set A * prob of selec prime no. from set B
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Set A consists of all the integers between 10 and 21, inclus 13 May 2024, 00:03
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ssinghal087

­Question:
In the question, y is also randomly selected which should meet the condition of y to be a prime number from set B. Therefore, along with the selection of x (i.e., multiple of 3 from set A), shouldn't there be selection of y as prime numbers from set B?

Probability of selecting multiple of 3 from set A * prob of selec prime no. from set B
­
I tried addressing that here:

https://gmatclub.com/forum/set-a-consis ... l#p3388778
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Set A consists of all the integers between 10 and 21, inclus 19 May 2025, 06:21
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