Total = 4*4 = 16
mn can be prime in 5 ways..
Probability is 5/16.
What is the OA.
Fun question
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beat_gmat_09
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set 1 = {1, 2, 3, 4}Viper83 wrote:If m is to be chosen at random from the set {1, 2, 3, 4} and n is to be chosen at random from the set {1, 3, 5, 7}, what is the probability that mn will be prime?
(A) 3/16
(B) 1/4
(C) 5/16
(D) 1/2
(E) 7/8
set 2 = {1, 3, 5, 7}
Probability that mn is prime -
Get probabilities for following -
1) 1 is selected from set 1 and 3,5,6 from set 2. = 1/4 * 3/4
2) 2 is selected from set 1 and 1 from set 2. = 1/4*1/4
3) 3 is selected from set 1 and 1 from set 2. = 1/4*1/4
3) 3 is selected from set 2 and 1 from set 1. = 1/4*1/4
4) 5 is selected from set 2 and 1 from set 1. = 1/4*1/4
5) 7 is selected from set 2 and 1 from set 1. = 1/4*1/4
Add all = 3/16 + 5/16 = 8/16= 1/2.
Option D.
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diebeatsthegmat
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.Viper83 wrote:If m is to be chosen at random from the set {1, 2, 3, 4} and n is to be chosen at random from the set {1, 3, 5, 7}, what is the probability that mn will be prime?
(A) 3/16
(B) 1/4
(C) 5/16
(D) 1/2
(E) 7/8
nm=14 ways
nm = prime so mn= (3*1, 2*1,5*1,7*1) =only 4 pairs
4/16=1/4
i dont know how you guys could find so many pairs of prime numbers!?!?!
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pesfunk
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Total 16 ways.
1 is neither prime nor composite, so the combinations possible for the question:
1,3
1,5
1,7
2,1
3,1
Hence the answer should be 5/16 or option C
What's the OA ?
1 is neither prime nor composite, so the combinations possible for the question:
1,3
1,5
1,7
2,1
3,1
Hence the answer should be 5/16 or option C
What's the OA ?
The Jock wrote:Total = 4*4 = 16
mn can be prime in 5 ways..
Probability is 5/16.
What is the OA.
Answer: C. (I agree with The Jock & pesfunk)
Approach:
a. note that 1 is not a prime.
b. from the given set, a prime can be obtained as a product only when you select "1" from one set & any prime from another set.
i.e., problem can be rewritten as "find probability of chosing one prime & "1" respectively from each set."
possible combinations: [1,2], [1,3], [1,5] etc. (refer the sets from "pesfunk")
hence result = 5/16.
Approach:
a. note that 1 is not a prime.
b. from the given set, a prime can be obtained as a product only when you select "1" from one set & any prime from another set.
i.e., problem can be rewritten as "find probability of chosing one prime & "1" respectively from each set."
possible combinations: [1,2], [1,3], [1,5] etc. (refer the sets from "pesfunk")
hence result = 5/16.
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beat_gmat_09
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set 1 = {1, 2, 3, 4}
set 2 = {1, 3, 5, 7}
possible combinations -
(1,2) - 1 from set 2 , 2 from set 1
(1,3) - 1 from set 2 , 3 from set 1
(3,1) - 3 from set 2 , 1 from set 1
(5,1) - 5 from set 2 , 1 from set 1
(7,1) - 7 from set 2 , 1 from set 1
(1,5) - 1 from set 1 , 5 from set 2
(1,7) - 1 from set 1 , 7 from set 2
(1,3) - 1 from set 1 , 3 from set 2
Am i going wrong anywhere ? please point out.
OA please.
set 2 = {1, 3, 5, 7}
possible combinations -
(1,2) - 1 from set 2 , 2 from set 1
(1,3) - 1 from set 2 , 3 from set 1
(3,1) - 3 from set 2 , 1 from set 1
(5,1) - 5 from set 2 , 1 from set 1
(7,1) - 7 from set 2 , 1 from set 1
(1,5) - 1 from set 1 , 5 from set 2
(1,7) - 1 from set 1 , 7 from set 2
(1,3) - 1 from set 1 , 3 from set 2
Am i going wrong anywhere ? please point out.
OA please.
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Ian Stewart
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You're counting several possibilities twice, those that I've highlighted (choosing, say, '3' from set two and '1' from set one is the same as choosing '1' from set one and '3' from set two). There are only 5 pairs which give a prime product, as a few people have listed above, so the answer is 5/16.beat_gmat_09 wrote:set 1 = {1, 2, 3, 4}
set 2 = {1, 3, 5, 7}
possible combinations -
(1,2) - 1 from set 2 , 2 from set 1
(1,3) - 1 from set 2 , 3 from set 1
(3,1) - 3 from set 2 , 1 from set 1
(5,1) - 5 from set 2 , 1 from set 1
(7,1) - 7 from set 2 , 1 from set 1
(1,5) - 1 from set 1 , 5 from set 2
(1,7) - 1 from set 1 , 7 from set 2
(1,3) - 1 from set 1 , 3 from set 2
Am i going wrong anywhere ? please point out.
OA please.
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beat_gmat_09
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Isn't this problem similar to rolling dice's problem, for example if two dice are rolled then probability of getting a sum ofIan Stewart wrote:You're counting several possibilities twice, those that I've highlighted (choosing, say, '3' from set two and '1' from set one is the same as choosing '1' from set one and '3' from set two). There are only 5 pairs which give a prime product, as a few people have listed above, so the answer is 5/16.beat_gmat_09 wrote:set 1 = {1, 2, 3, 4}
set 2 = {1, 3, 5, 7}
possible combinations -
(1,2) - 1 from set 2 , 2 from set 1
(1,3) - 1 from set 2 , 3 from set 1
(3,1) - 3 from set 2 , 1 from set 1
(5,1) - 5 from set 2 , 1 from set 1
(7,1) - 7 from set 2 , 1 from set 1
(1,5) - 1 from set 1 , 5 from set 2
(1,7) - 1 from set 1 , 7 from set 2
(1,3) - 1 from set 1 , 3 from set 2
Am i going wrong anywhere ? please point out.
OA please.
12, where two same sets are considered, so possibilities for getting 12 of sum involves (7,5) and (5,7) as different possibilities and both the possibilities should be counted. I counted (1,3) and (1,3) two times but for others ex (1,7) ; (7,1) i thought of the dice problem.
Please correct me if i am wrong.
Thanks
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When you're counting (5,7) twice in a dice problem (I guess you're imagining dice labeled from 2 to 7 or something), you're counting them twice because you might get 5 on the first die and 7 on the second, or 7 on the first die and 5 on the second. You would not count, for example, (7,7) twice, because there's only one way for that to happen: you must get 7 on each die.beat_gmat_09 wrote: Isn't this problem similar to rolling dice's problem, for example if two dice are rolled then probability of getting a sum of
12, where two same sets are considered, so possibilities for getting 12 of sum involves (7,5) and (5,7) as different possibilities and both the possibilities should be counted. I counted (1,3) and (1,3) two times but for others ex (1,7) ; (7,1) i thought of the dice problem.
In the problem in the original post, there's only one way to get the set (1,7); you must pick '1' from the first set, and '7' from the second set. You don't want to count that twice.
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beat_gmat_09
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Thanks Ian but i'm not convinced.Ian Stewart wrote: When you're counting (5,7) twice in a dice problem (I guess you're imagining dice labeled from 2 to 7 or something), you're counting them twice because you might get 5 on the first die and 7 on the second, or 7 on the first die and 5 on the second. You would not count, for example, (7,7) twice, because there's only one way for that to happen: you must get 7 on each die.
In the problem in the original post, there's only one way to get the set (1,7); you must pick '1' from the first set, and '7' from the second set. You don't want to count that twice.
I think it is same as the dice problem and the 2 ways of selecting should be counted.
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Well try this question using your method:beat_gmat_09 wrote:Thanks Ian but i'm not convinced.Ian Stewart wrote: When you're counting (5,7) twice in a dice problem (I guess you're imagining dice labeled from 2 to 7 or something), you're counting them twice because you might get 5 on the first die and 7 on the second, or 7 on the first die and 5 on the second. You would not count, for example, (7,7) twice, because there's only one way for that to happen: you must get 7 on each die.
In the problem in the original post, there's only one way to get the set (1,7); you must pick '1' from the first set, and '7' from the second set. You don't want to count that twice.
I think it is same as the dice problem and the 2 ways of selecting should be counted.
Choose one number from the set {1, 2} and one number from the set {3, 4}. What is the probability the product is even? Using your method, you'll get an answer greater than 100%, which can't be right.
It is true that this is similar to a dice problem, but when we count say 6,5 and 5,6 in a dice problem as two ways to get a sum of 11, we count this twice because in the first case, we get the 6 on the first die, and in the second case we get the 5 on the first die. We aren't double-counting anything. In the question above, if you're going to get 1, 7, there's only one way to do it, because you can't pick the '7' from the first set; it isn't there.
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beat_gmat_09
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Thanks, got itIan Stewart wrote: It is true that this is similar to a dice problem, but when we count say 6,5 and 5,6 in a dice problem as two ways to get a sum of 11, we count this twice because in the first case, we get the 6 on the first die, and in the second case we get the 5 on the first die. We aren't double-counting anything. In the question above, if you're going to get 1, 7, there's only one way to do it, because you can't pick the '7' from the first set; it isn't there.
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