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Set A consists of the values 1, 2, and 3. Set B consists of the values

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Set A consists of the values 1, 2, and 3. Set B consists of the values  [#permalink]

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New post 11 Oct 2018, 01:50
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Question Stats:

65% (02:25) correct 35% (02:45) wrong based on 65 sessions

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Set A consists of the values 1, 2, and 3. Set B consists of the values 5, 6, and 7. One number is selected at random from each set. The two selected numbers are then added together.

The probability that the sum is even is how much greater than the probability that the sum is a prime number?


A. \(\frac{1}{9}\)

B. \(\frac{2}{9}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{5}\)

E. \(\frac{3}{5}\)

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Set A consists of the values 1, 2, and 3. Set B consists of the values  [#permalink]

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New post Updated on: 11 Oct 2018, 05:08
Bunuel wrote:
Set A consists of the values 1, 2, and 3. Set B consists of the values 5, 6, and 7. One number is selected at random from each set. The two selected numbers are then added together.

The probability that the sum is even is how much greater than the probability that the sum is a prime number?


A. \(\frac{1}{9}\)

B. \(\frac{2}{9}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{5}\)

E. \(\frac{3}{5}\)


Probability of Sum to be even Number = First odd second Odd OR first even second even = (2/3)*(2/3) + (1/3)*(2/3) = 5/9
OR
Probability of Sum to be odd Number = First odd second even OR first even second odd = (2/3)*(1/3) + (1/3)*(2/3) = 4/9
i.e. Probability of Sum to be even = 1-(4/9) = 5/9


Probability of Sum to be Prime Number = Probability of (1,6) or (2,5) = (1/3)*(1/3) + (1/3)*(1/3) = 2/9

Difference required = (5/9) - (2/9) = 3/9 = 1/3

Answer: Option C
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Originally posted by GMATinsight on 11 Oct 2018, 03:02.
Last edited by GMATinsight on 11 Oct 2018, 05:08, edited 1 time in total.
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Re: Set A consists of the values 1, 2, and 3. Set B consists of the values  [#permalink]

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New post 11 Oct 2018, 03:48
GMATinsight:
Taking your solution:

1. Probability of Sum to be even Number = First odd second even OR first even second odd = (2/3)*(1/3) + (1/3)*(2/3) = 4/9

In 1 it seems you have calculated the probability of getting an odd number, for even number of the set the probability should be

(Odd+ odd), (even+ even) i.e
( 2/3*2/3)+(1/3*1/3)= 5/9

Probability of Sum to be Prime Number = Probability of (1,6) or (2,5) = (1/3)*(1/3) + (1/3)*(1/3) = 2/9

Difference required = (5/9) - (2/9) = 3/9= 1/3



Answer: Option C
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Re: Set A consists of the values 1, 2, and 3. Set B consists of the values  [#permalink]

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New post 11 Oct 2018, 03:58
1. Probability of Sum to be even Number = First odd second even OR first even second odd = (2/3)*(1/3) + (1/3)*(2/3) = 4/9

In 1 it seems you have calculated the probability of getting an odd number, for even number of the set the probability should be

(Odd+ odd), (even+ even) i.e
( 2/3*2/3)+(1/3*1/3)= 5/9

Probability of Sum to be Prime Number = Probability of (1,6) or (2,5) = (1/3)*(1/3) + (1/3)*(1/3) = 2/9

Difference required = (5/9) - (2/9) = 3/9= 1/3



Answer: Option C[/quote]



Should'nt the Probability of sum of prime numbers be multiplied by 2 i.e. 4/9. So the final answer be 5/9-4/9 = 1/9 Answer(A)
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Re: Set A consists of the values 1, 2, and 3. Set B consists of the values  [#permalink]

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New post 11 Oct 2018, 05:11
audi006 wrote:
1. Probability of Sum to be even Number = First odd second even OR first even second odd = (2/3)*(1/3) + (1/3)*(2/3) = 4/9

In 1 it seems you have calculated the probability of getting an odd number, for even number of the set the probability should be

(Odd+ odd), (even+ even) i.e
( 2/3*2/3)+(1/3*1/3)= 5/9

Probability of Sum to be Prime Number = Probability of (1,6) or (2,5) = (1/3)*(1/3) + (1/3)*(1/3) = 2/9

Difference required = (5/9) - (2/9) = 3/9= 1/3



Answer: Option C




Should'nt the Probability of sum of prime numbers be multiplied by 2 i.e. 4/9. So the final answer be 5/9-4/9 = 1/9 Answer(A)[/quote]

Answering the highlighted part

There is no need to multiply the probability by 2 because first number can't be 6 and second can't be 1. Here the order is fixed that first set must give 1 and second must give 6 for sum to be 7
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Re: Set A consists of the values 1, 2, and 3. Set B consists of the values   [#permalink] 11 Oct 2018, 05:11
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