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

Question

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}

GMATinsight · Answer

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

Archit3110 · Answer

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

audi006 · Answer

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)

GMATinsight · Answer

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)

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

santosh93 · Answer

I know the basic formula for probability i.e.

p = \frac{Favourable cases}{ Total Number cases}

Given two set A = {1,2,3} B= {5,6,7}.

So ordered pairs formed by choosing one from each set are
(1,5)(1,6)(1,7)
(2,5)(2,6)(2,7)
(3,5)(3,6)(3,7) or simply we can get this by multiplying the number of elements in each set 
Therefore the total number of cases = 9

Ordered pair whose sum is even are (1,5) (1,7)(2,6)(3,5)(3,7) = 5 
Ordered pair whose sum is prime are (1,6) (2,5) = 2

Let 
 P1 = Probality of sum is even =  \frac{5}{9} 
 P2 = Probality of sum is prime =  \frac{2}{9}

The probability that the sum is even(P1) is  how much greater  than the probability that a sum is a prime number(P2)
= P1 - P2
= \frac{5}{9}  -  \frac{2}{9} 
= \frac{3}{9} 
= \frac{1}{3}

Hence the answer is choice  C

PS:   I got this question wrong as I misread how  much  greater to How  many times  greater . Wish No one does this error in GMAT test

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

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