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M08-05

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M08-05  [#permalink]

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New post 15 Sep 2014, 23:36
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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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Re M08-05  [#permalink]

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New post 15 Sep 2014, 23:36
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: M08-05  [#permalink]

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New post 23 Aug 2015, 22:12
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: M08-05  [#permalink]

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New post 23 Aug 2015, 22:14
srivelivala wrote:
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


You don't have a 3 in set B.
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Re: M08-05  [#permalink]

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New post 18 Oct 2015, 02:41
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: M08-05  [#permalink]

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New post 18 Oct 2015, 10:18
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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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

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Re: M08-05  [#permalink]

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New post 18 May 2017, 05:26
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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Re: M08-05  [#permalink]

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New post 18 May 2017, 06:24
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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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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Re: M08-05  [#permalink]

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New post 04 Jul 2017, 06:53
1
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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Re: M08-05  [#permalink]

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New post 06 Oct 2018, 04:11
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 ?
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Re: M08-05 &nbs [#permalink] 06 Oct 2018, 04:11
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