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Re: M08-05 [#permalink]
Bunuel wrote:
Official Solution:


Two sets are defined as follows:

\(A = \{2, 3, 4, 4, 4\}\)

\(B = \{0, 1, 2\}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?


A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)


The total # of outcomes is \(5*3=15\).

Now, as set \(B\) is smaller, then work with it:

0 can be added to 2 or to 3 to get a prime: \(1+1=2\) cases;

1 can be added to 2 or to either of three 4's to get a prime: \(1+3=4\) cases;

2 can be added only to 3 to get a prime: 1 case.

So, the total # of favorable outcomes is \(2+4+1=7\). \(P=\frac{\text{favorable}}{\text{total}}=\frac{7}{15}\).


Answer: D



How come there are 15 total possible outcomes... when there are only 3 distinct elements in set A.
I believe the possible outcomes should only be 9.
plz correct me if im wrong
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Re: M08-05 [#permalink]
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jamesav wrote:
Bunuel wrote:
Official Solution:


Two sets are defined as follows:

\(A = \{2, 3, 4, 4, 4\}\)

\(B = \{0, 1, 2\}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?


A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)


The total # of outcomes is \(5*3=15\).

Now, as set \(B\) is smaller, then work with it:

0 can be added to 2 or to 3 to get a prime: \(1+1=2\) cases;

1 can be added to 2 or to either of three 4's to get a prime: \(1+3=4\) cases;

2 can be added only to 3 to get a prime: 1 case.

So, the total # of favorable outcomes is \(2+4+1=7\). \(P=\frac{\text{favorable}}{\text{total}}=\frac{7}{15}\).


Answer: D



How come there are 15 total possible outcomes... when there are only 3 distinct elements in set A.
I believe the possible outcomes should only be 9.
plz correct me if im wrong


Number 4 is in set A three times. So, the probability of picking 4 is higher, than the probability of picking 2 or 3, which should be taken into account.
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Re: M08-05 [#permalink]
Bunuel wrote:
Two sets are defined as follows:

\(A = \{2, 3, 4, 4, 4\}\)

\(B = \{0, 1, 2\}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?


A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)




KarishmaB can you pls explain why arent we considering cases like (0;2) and (2:0) as separate ones and (2;1) and (1;2) and other similar cases
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Re: M08-05 [#permalink]
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dave13 wrote:
Bunuel wrote:
Two sets are defined as follows:

\(A = \{2, 3, 4, 4, 4\}\)

\(B = \{0, 1, 2\}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?


A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)




KarishmaB can you pls explain why arent we considering cases like (0;2) and (2:0) as separate ones and (2;1) and (1;2) and other similar cases


A case of (2, 0) is possible when we take 2 from set A and 0 from set B.
How is the case of (0, 2) possible? There is no 0 in set A. So in case of (0, 2), 0 will be taken from set B only and 2 from set A only. Then this is the same as (2, 0) discussed above and hence cannot be counted twice.
Whether you pick the number from set A first or from set B first is irrelevant because we obtain only one case. There is no arrangement here. We have to just select 1 number from A and 1 number from B.
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Re: M08-05 [#permalink]
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M08-05 [#permalink]
­Total number of possible outcomes: 5 * 3 = 15

Min Sum = 2 + 0 = 2
Max Sum = 4 + 2 = 6

Prime numbers in the range 2-6, inclusive are 2, 3, 5

        Group A                Group B
2      2                          0
3      3                          0
        2                          1

5      4 (* 3 times)          1
        3                          2


=> There are total 7 combinations

=> Probability = \(\frac{7}{15}\)
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