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# Set A consists of all the integers between 10 and 21, inclus

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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
GyanOne wrote:
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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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
pooja.sharma2909 wrote:
GyanOne wrote:
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.

Hope it helps.
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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]

Bunuel wrote:
pooja.sharma2909 wrote:
GyanOne wrote:
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.

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?
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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Contropositive wrote:
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?

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 [#permalink]
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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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]

hughng92 wrote:
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?

Posted from my mobile device

­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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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
Bunuel wrote:
pooja.sharma2909 wrote:
GyanOne wrote:
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.

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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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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ssinghal087 wrote:

­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

­