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Two integers will be randomly selected from the sets above [#permalink]
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17 Jan 2008, 14:50
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A = {2, 3, 4, 5} B = {4, 5, 6, 7, 8} Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ? (A) 0.15 (B) 0.20 (C) 0.25 (D) 0.30 (E) 0.33 OPEN DISCUSSION OF THIS QUESTION IS HERE: twointegerswillberandomlyselectedfromthesetsabove143451.html
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Last edited by Bunuel on 06 Jul 2013, 08:28, edited 1 time in total.
Edited the question.



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Re: PS  probability [#permalink]
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17 Jan 2008, 14:59
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B\(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\) or \(p=\frac{4}{4}*\frac{1}{5}=0.20\) or \(p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20\)
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Re: PS  probability [#permalink]
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You don't need to perform any calculations in this example. Regardless of the number you choose from set A, there is only one number (out of 5) that makes the sum equal 9. Therefore, the probability is 1/5. The answer is B.



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Re: PS  probability [#permalink]
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B total 20 ways to sum, 4 of the sums equals 9 so, prob is 4/20= 1/5= .20



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Re: PS  probability [#permalink]
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24 Aug 2008, 14:52
vermatanya wrote: Quote: A = {2,3,4,5} B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33
thanx
p= 4/(4c1*5c1)=1/5=0.2
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Re: PS  probability [#permalink]
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07 Sep 2009, 11:29
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Maybe I am late with response, but anyway. We have two sets with 4 and 5 integers included in each one respectively. Thus, we have 4*5=20 possible ways of selecting two integers from these sets. The requirement is that we need the sum of 9. Here we have 4 possible choices: 2 and 7 3 and 6 4 and 5 5 and 4 Probability = favourable / total 4/20 = 0.2
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Re: PS  probability [#permalink]
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27 Sep 2009, 06:17
A = {2,3,4,5} B = {4,5,6,7,8}
Total number of possibilities of choosing two numbers is = 4 * 5 = 20 ways
No. of possibilities of choosing numbers such that their sum is 9 is: (2,7),(3,6),(4,5),(5,4) thus 4 ways
So probability is = 4/20 = 1/5 = 0.2



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Re: PS  probability [#permalink]
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14 Feb 2010, 13:24
vermatanya wrote: A = {2,3,4,5} B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33
thanx
Total ways of choosing 1 number from Set A and 1 from Set B.... 4c1 x 5c1 = 20 Comb which yield 9 as the sum  2,7.....3,6.....4,5....5,4 = 4 comb... Probability = 4 / 20 = .20 = B
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Re: PS  probability [#permalink]
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15 Feb 2010, 00:24
vermatanya wrote: Quote: A = {2,3,4,5} B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33
thanx
P(A) = 1 P(B) = 1/5 P(A&B) = 1/5 = 0.2 hence B



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Re: PS  probability [#permalink]
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24 Feb 2012, 21:54
walker wrote: B
\(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\)
or
\(p=\frac{4}{4}*\frac{1}{5}=0.20\)
or
\(p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20\) Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?



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Re: PS  probability [#permalink]
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fortsill wrote: walker wrote: B
\(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\)
or
\(p=\frac{4}{4}*\frac{1}{5}=0.20\)
or
\(p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20\) Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? I'd offer another approach which I hope will also clarify the above. A = {2,3,4,5} B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 Rearrange the first set: A = {5, 4, 3, 2}B = {4, 5, 6, 7, 8}As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2. Answer: B. Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5. Answer: B. Hope it's clear.
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Re: PS  probability [#permalink]
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03 May 2012, 05:11
Bunuel wrote: fortsill wrote: walker wrote: B
\(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\)
or
\(p=\frac{4}{4}*\frac{1}{5}=0.20\)
or
\(p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20\) Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? I'd offer another approach which I hope will also clarify the above. A = {2,3,4,5} B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 Rearrange the first set: A = {5, 4, 3, 2}B = {4, 5, 6, 7, 8}As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2. Answer: B. Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5. Answer: B. Hope it's clear. Bunuel, I am confused. We are selecting a pair from a total of 9 numbers or 7 numbers (if we eliminate duplicates). So we will be getting for total number of ways, in either case 2 from 9 is 36 and 2 from 7 is 21 So we have [ (2, 7),(3,6), (4,5) ] 3 pairs or [ (2, 7),(3,6), (4,5) (5,4) ] 4 pairs if we don't eliminate duplicates Either way, I am getting the wrong answer 3/36 and 4/21 are both wrong. Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same. How is it different from this one (the general concept)? iftwoofthefourexpressionsxyx5yxy5xyare92727.html#p713823Thank you



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Re: PS  probability [#permalink]
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03 May 2012, 09:29
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Lstadt wrote: 2 from 9 is 36 and 2 from 7 is 21
You are not selecting 2 numbers from 9 numbers. You are selecting 1 number from 4 numbers and 1 number from 5 numbers. Mind you, they are different. While selecting 2 numbers from {1, 2, 3, 4, 5, 6, 7, 8, 9}, you can select (3, 4) While selecting 1 from 4 and 1 from 5: {1, 2, 3, 4} and {5, 6, 7, 8, 9}, you cannot select (3, 4). That's the problem number 1. Lstadt wrote: Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same.
Secondly, the two given sets are distinct: A = {2,3,4,5} B = {4,5,6,7,8} When we pick a number from A and from B, we get {4, 5} or we can do it in this way {5, 4}. These are two different cases. In the first case, the 4 comes from A and in the second case, it comes from B.
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Re: 2 integers will be randomly selected from the sets above [#permalink]
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20 Aug 2012, 04:12
Responding to a pm: No, there will not be 8 different pairs. A = {2,3,4,5} B = {4,5,6,7,8} If you select 2 from A, from B you must select 7. If you select 3 from A, from B you must select 6. If you select 4 from A, from B you must select 5. If you select 5 from A, from B you must select 4. The total number of ways in which the sum can be 9 is 4. The question was whether the last two cases are distinct or not. Is 4 from A and 5 from B the same as 4 from B and 5 from A. Since 4 and 5 come from different sets in the two cases, they represent two different cases and should be counted as such.
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Re: 2 integers will be randomly selected from the sets above [#permalink]
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15 Nov 2012, 03:25
HI Karishma,
I understood the question, got 0.20, but I have a doubt.
At first, I got 6 pairs  (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.
How do we come too know that we need not include duplicate pairs??
I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q?



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Re: 2 integers will be randomly selected from the sets above [#permalink]
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15 Nov 2012, 03:33
aditi2013 wrote: HI Karishma,
I understood the question, got 0.20, but I have a doubt.
At first, I got 6 pairs  (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.
How do we come too know that we need not include duplicate pairs??
I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q? You need to select a pair {a, b} such that their sum is 9. a is selected from set A and b is selected from set B. a can take a value from set A only i.e. one of 2/3/4/5 b can take a value from set B only i.e. one of 4/5/6/7/8 How do you select {6, 3}? a cannot be 6. A selection is different from another when you select different numbers from each set. One selection is {4, 5}  4 from A, 5 from B Another selection is {5, 4}  5 from A, 4 from B One selection is {3, 6}  3 from A, 6 from B There is no such selection {6, 3}  A has no 6.
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Re: 2 integers will be randomly selected from the sets above [#permalink]
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15 Nov 2012, 03:40
when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order??



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Re: 2 integers will be randomly selected from the sets above [#permalink]
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15 Nov 2012, 03:46
aditi2013 wrote: when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order?? It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter. {4, 5}  4 is from A and 5 is from B {5, 4}  5 is from A and 4 is from B These two are different. {2, 6}  2 is from A and 6 is from B {6, 2}  2 is from A and 6 is from B These two are same.
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Re: PS  probability [#permalink]
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06 Jul 2013, 08:22
walker wrote: B
\(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\)
or
\(p=\frac{4}{4}*\frac{1}{5}=0.20\)
or
\(p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20\) I made it right, but I don´t understand this approach (Combinatorics). Why the denominator in the first solution \(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\) is \(\ C^4_1\) Some help? Thanks!
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Re: PS  probability [#permalink]
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06 Jul 2013, 08:36
Maxirosario2012 wrote: walker wrote: B
\(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\)
or
\(p=\frac{4}{4}*\frac{1}{5}=0.20\)
or
\(p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20\) I made it right, but I don´t understand this approach (Combinatorics). Why the denominator in the first solution \(p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20\) is \(\ C^4_1\) Some help? Thanks! A = {2, 3, 4, 5} B = {4, 5, 6, 7, 8}
Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9 ?(A) 0.15 (B) 0.20 (C) 0.25 (D) 0.30 (E) 0.33 The probability = (# favorable outcomes)/(total # of outcomes) (# favorable outcomes) is 4C1=4: (2, 7); (3, 6), (4, 5) and (5, 4). (total # of outcomes) is 4C1*5C1=4*5=20. P=4/20. Answer: B. Also chekc here: twointegerswillberandomlyselectedfromthesetsabove58629.html#p1049709OPEN DISCUSSION OF THIS QUESTION IS HERE: twointegerswillberandomlyselectedfromthesetsabove143451.html
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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