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Probability problem need a better understanding (m08q05) [#permalink]
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05 Jul 2008, 05:30
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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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) \(\frac{9}{15}\) Source: GMAT Club Tests  hardest GMAT questions The total outcomes for the problem is 15. However when we calculate the sum of 2 sets that generate a prime number, Do we count 4 once or three times. In other words, 2+0 2+1 3+0 3+2 4+1 or 2+0 2+1 3+0 3+2 4+1 4+1 4+1 If so why?



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Re: Probability problem need a better understanding [#permalink]
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05 Jul 2008, 06:19
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Quote: The total outcomes for the problem is 15. However when we calculate the sum of 2 sets that generate a prime number, Do we count 4 once or three times. If you have counted the number of all outcomes as 15 (5*3), you have included the pairs with 4 more than one time. So, when counting the number of favourable outcomes, you need again to count pairs with 4 more than one time. N(fav) = 2 + 2 + 1+1+1 =7. P=N(fav)/N(total) = 7/15. Alternative approach: P(select 2) = 1/5; and 2 of 3 numbers from the second set will give us prime – 1/5*2/3 to get prime P(select 3) = 1/5; and 2 of 3 numbers from the second set will give us prime – 1/5*2/3 to get prime P(select 4) = 3/5; and 1 of 3 numbers from the second set will give us prime – 3/5*1/3 to get prime P = 2/15+2/15+3/15 = 7/15.



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Re: Probability problem need a better understanding [#permalink]
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06 Jul 2008, 08:27
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another way to look at it is...
these are balls in a bag with numbers printed on them.
As there are more 4s as compared to 3s or 2s, we need to count 4 more than one time, as chances of picking a 4 numbered ball are more.



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Re: Probability problem need a better understanding (m08q05) [#permalink]
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15 Jan 2010, 07:21
D. But I did it the long way. 5 numbers in the first set and 3 in the second. Yields 15 total possibilities. Then I counted each one that could be prime. Got 7.



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15 Jan 2010, 08:18
7/15. But I counted each combination of prime numbers and came up with the answer. For large problems, we might need to use the probability of individual combination as suggested few posts above. Suresh.



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Re: Probability problem need a better understanding (m08q05) [#permalink]
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15 Jan 2010, 21:15
total possible combination : 5C1*3C1=15 favorable out com:2+0,2+1,3+0 ,3+2,4+1,4+1,4+1 so P=7/15
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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17 Jan 2010, 08:28
probability = p(e)/p(n)
p(n)=total no of of possible outcomes = 5c1*3c1=15
I counted all the possibilities for prime numbers
2+0 2+1 3+0 3+2 4+1 4+1 4+1
so there are 7 possibilities
so p(n)=7/15
so my ans is d



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Re: Probability problem need a better understanding (m08q05) [#permalink]
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24 Jan 2010, 01:05
very easy question  you count 4+1 three times because there are three distinct 4's in set A 2+0 2+1 3+0 3+2 4+1 4+1 4+1 So 7/(5*3) = 7/15 =>D
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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20 Jan 2011, 07:10
D.
Following combinations will give the desired result: 2 0 3 0 2 1 4 1 (multiply by 3... as 4 occurs 3 times) 3 2 Total = 7
Total possibilities = 5C1 * 3C1 = 15
Probability = 7/15



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20 Jan 2011, 07:16
Same here,
Counted all 15 possibilities, No doubt we need to include 4 thrice as it is asking for 'Probability' not 'Permutation & Combination'. Thats the trick ..
In P&C we need to ignore repeating objects but not in 'Probability'
Answer is 7/15



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Re: Probability problem need a better understanding (m08q05) [#permalink]
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20 Jan 2011, 07:33
Good post. I got it to C and D and had the same inquiry. Thanks for the replies :D
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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21 Jan 2011, 04:56
I have done by long method When we will take 0 from the B set and match with numbers in set A; only 2, and 3 from the A set can give a prime number. So, probability of picking 2, and 3 is 2/5 and of 0 is 1/3. This gives combined probability of 2/5*1/3=2/15 (Combined probability will decrease so I have used multiplication rule). Next, I have taken 1 from the set B and it can give a prime number when combined with 2,4,4, and 4. So using the same reasoning above its combined probability will be 4/5*1/3=4/15. In last, 2 from the set B can give only prime number when combined with 1 from the set A. So its combined probability will be1/5*1/3=1/15.
All the 3 probability I have added to get the final answer because probability should increase 2/15+4/15+1/15= 7/15.
I think my answer and approach are correct but can somebody will tell me how to solve this problem within 1 minute and I often use to do mistake in counting the number of probable case How to avoid this.
Cheers!!!



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Re: Probability problem need a better understanding (m08q05) [#permalink]
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21 Jan 2011, 05:31
rajeshaaidu wrote: I think my answer and approach are correct but can somebody will tell me how to solve this problem within 1 minute and I often use to do mistake in counting the number of probable case How to avoid this.
Cheers!!! Your method shouldn't take longer than 1 min? Since we only have 3 different integers, I find that calculating the propability for each number is the most logical way for me. 2 is a prime when +0 and +1 => We draw 2 from the first set 1/5 and we draw 0 or 1 from the second 2/3 => 1/5*2/3 = 2/15 3 is a prime when +0 and +2 => We draw 3 from the first set 1/5 and we draw 0 or 2 from the second 2/3 => 1/5*2/3 = 2/15 4 is a prime when +1 => We draw 4 from the first set 3/5 and we draw 1 from the second set 1/3 = 3/5*1/3 = 3/15 2+2+3 = 7/15 This method shouldnt take more than one minute
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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24 Jan 2012, 08:45
picked one digit from set one and added with every digits separately from set 2 to get a prime, and also got the total to get the final probability.
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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24 Jan 2012, 11:41
rajeshaaidu wrote: I think my answer and approach are correct but can somebody will tell me how to solve this problem within 1 minute and I often use to do mistake in counting the number of probable case How to avoid this. 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) \(\frac{9}{15}\) Total # of outcomes 5*3=15. Now, as set B is smaller 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 cases. So total # of favorable outcomes is 2+4+1=7. P=favorable/total=7/15. Answer: D.
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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24 Jan 2012, 18:49
We can use the counting method in this case since the number of favourable cases are not many.
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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24 Jan 2013, 06:28
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To solve such questions, one should keep in mind that the testmakers design such questions to be solved in two mins or less even if one chooses to LIST all possible one by one. The most important thing is to quickly decide the method to use, then LIST away. In this particular question, there are only 7 desirable outcomes out of a total of 15 possible outcomes. Surely, it would take one about thirty to fortyfive seconds to list all 7 prime outcomes. Total possible outcomes= 5C1 * 3C1 (i.e. picking one item each from two sets of 5 and 3 items respectively= 5*3= 15 Desired prime outcomes from listing= 2,3,3,5,5,5,5 =7 Probability= Desired prime outcomes/ Total possible outcomes = 7/15 Cheers, Der alte Fritz.
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Re: Probability problem need a better understanding (m08q05) [#permalink]
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24 Jan 2013, 07:22
The required probability = No. of outcomes when the sum comes to be a prime / total possible outcomes No. of outcomes when the sum comes to be a prime: 2+0 2+1 3+0 3+2 4+1 4+1 4+1 i.e 7 required outcomes Total possible outcomes = 5 x 3=15 Therefore, The required probability= 7/15
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