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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. \(\frac{9}{15}\)
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16 Sep 2014, 00:36
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
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Re: M0805
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23 Aug 2015, 23:12
Can someone please explain why we do not consider (2,3) & (3,2) as two separate possible outcomes and count them both while calculating the probability, similar to what we do when we roll two dice and calculate, the sum of both dice to be say 8. Thanks, Srinivas



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Re: M0805
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18 Oct 2015, 03:41
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: M0805
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18 Oct 2015, 11:18
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: M0805
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18 May 2017, 06:26
favourable outcome are (2,0) (2,1) (3,0) (3,2) (4,1) (4,1) (4,1) hence probability is 7/15 but i considered all (4,1) is similar then my total favourable outcome is 5. hence i marked ans 5/15.
plz clarify.



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18 May 2017, 07:24
digvijay99 wrote: favourable outcome are (2,0) (2,1) (3,0) (3,2) (4,1) (4,1) (4,1) hence probability is 7/15 but i considered all (4,1) is similar then my total favourable outcome is 5. hence i marked ans 5/15.
plz clarify. Since there are three 4's in A, then the probability of getting (4, 1) is greater than if there were just one 4.
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Re: M0805
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04 Jul 2017, 07:53
Hi Bunuel,
Only elements of type {A,B} are considered or {B,A} are considered. Why not both? Type {A,B}: (2,0) (2,1) (3,0) (3,2) (4,1) (4,1) (4,1) Type {B,A} (0,2)(0,3)..like this we will again get here 7 possible outcomes. I didn't understand why we haven't considered both types of sets. Could you please explain?



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06 Oct 2018, 05:11
Bunuel, I got the answer correct at first place but I used guess. What I was trying before I guess is : P = X/N so N = Total number of ways to select 2 numbers out of 8 is 2C8 and X= Ways to pair prime number and select 2 out of it so it is 2C7 and after calculations it gives me 3/4. Could you please help where I faltered ?










