M10-14

Question

Set  S  consists of all prime integers less than 10. If two numbers are drawn at random one after the other, with replacement, from this set, what is the probability that the sum of these two numbers is odd?

A.  \frac{1}{8}  
B.  \frac{1}{6}  
C.  \frac{3}{8}  
D.  \frac{1}{2}  
E.  \frac{5}{8}

Bunuel · Answer

Official Solution:

Set  S  consists of all prime integers less than 10. If two numbers are drawn at random one after the other, with replacement, from this set, what is the probability that the sum of these two numbers is odd?

A.  \frac{1}{8}  
B.  \frac{1}{6}  
C.  \frac{3}{8}  
D.  \frac{1}{2}  
E.  \frac{5}{8}

Set  S =\{2, 3, 5, 7\} . Since the sum of the numbers must be odd, then one of the numbers must even prime, so 2, and another must be any odd prime.
 
 P(	ext{sum is odd}) =  
 
 =P(	ext{first number is 2, another is any odd prime}) + P(	ext{first number is any odd prime, another is 2})=   
 
 = \frac{1}{4}*\frac{3}{4} + \frac{3}{4}*\frac{1}{4} = \frac{6}{16} = \frac{3}{8} .

Answer: C

rsamant · Answer

I quickly listed out the set of numbers (2,2) (2,3) (2,5) (2,7) (3,3) (3,5) (3,7) (5,5) (5,7) and (7,7) and then I counted the only odd pair which left me with a probability of 3/10. Why is this method incorrect?

Bunuel · Answer

There are more possibilities when picking two numbers:

(2,2)
 (2,3) 
(3,2)
(2,5)
(5,2) 
(2,7)
(7,2)  
(3,3) 
(3,5)
(5,3) 
(3,7)
(7,3) 
(5,5) 
(5,7)
(7,5)
(7,7)

P = 6/16 = 3/8.

varmashreekanth · Answer

Hi Bunuel,

Does this 'Not necessarily different' terminology given in the question implies that the number is not repeated?
Because when I first attempted this question i got 1/2 i.e . (1/4)*(3/3) + (3/4)*(1/3) = (1/2)

But now when I saw your solution, I came to know that the numbers should not be reduced and should be kept same.
I just wanted to get it confirmed with you that everytime in future if I see this term 'Not necessarily different' should I consider that the number is not repeated?

Bunuel · Answer

If a number is selected from set S at random and then another number,  not necessarily different , is selected from set S at random, ...

The highlighted part in the question means that from S={2,3,5,7} we can choose any number more than once. So, we can choose the following two numbers:
 (2,2) 
 (2,3) 
(3,2)
(2,5)
(5,2) 
(2,7)
(7,2)  
 (3,3)  
(3,5)
(5,3) 
(3,7)
(7,3) 
 (5,5)  
(5,7)
(7,5)
 (7,7)

jayditya · Answer

why 1 is not considered as a prime number?

Bunuel · Answer

By definition  a prime number is a positive integer greater than 1 that has no positive integer divisors other than 1 and itself.

The main reason 1 is not conceded as prime is because of  the fundamental theorem of arithmetic  (unique prime factorization theorem), which states that  every integer greater than 1 can be represented  uniquely  as a product of prime numbers .

For example,  60 = 2^2*3*5  (the order of the primes does not matter here). Now, if we allow 1 to be a prime then we can represent 60 as  1*2^2*3*5  or as  1^2*2^2*3*5  or as  1^3*2^2*3*5  as  1^4*2^2*3*5  ... As you can see the representation is no longer unique and thus the fundamental theorem of arithmetic is no longer correct, which is a problem because the  fundamental  theorem of arithmetic is called  fundamental  for a reason.

2. Properties of Integers  
     Theory   
 Math: Number Theory 
 Number Properties: Tips and hints 
 Advanced Number Properties on GMAT (All 5 parts) 
 Do you make these 3 mistakes in GMAT Even-Odd Questions?  
 Everything about Factorials on the GMAT 
 Solving Complex Problems Using Number Line 
     Questions   
 Tough New Set: Number Properties Set 
 Power of a Number in a Factorial Problems 
 Number Properties Ability Quiz 
 Special Numbers and Sequences Problems 
 The E-GMAT Number Properties Knockout: 10 Great 700+ Ques for you! 
 Trailing Zeros Problems 
 DS Questions 
 PS Questions

For other subjects:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

phamminh2003 · Answer

I find 4P1*3P1 = 12 when I calculate the total times that two numbers are drawn at random one after the other, with replacement. Then, I find 3 pairs of number (2;3),(2;5),(2;7),(3;2),(5;2),(7;2) that the sum is odd 
=> probability : 6/12=1/2
What is my mistake here?

Bunuel · Answer

When picking with replacement, we can get 16 pairs, not 12. The numbers can repeat:

(2,2)
 (2,3)
(3,2)
(2,5)
(5,2)
(2,7) 
(7,2)
(3,3)
(3,5)
(5,3)
(3,7)
(7,3)
(5,5)
(5,7)
(7,5)
(7,7)

P = 6/16 = 3/8.

ItzSam · Answer

Since question mentions integers, then why can't the set have -2, -3, -5 etc. as the values? These are also less than 10.

Bunuel · Answer

We are told that Set S consists of all  prime  integers less than 10. Only positive integers can be primes, with 2 being the smallest prime number.

HowCanIDoThis · Answer

I like the solution - it’s helpful.

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