Set S is the prime integers between 0 and 20. If three numbers are cho

Question

Set S is the prime integers between 0 and 20. If three numbers are chosen randomly from set S, what is the probability that the sum of these three numbers is odd?

(A) 15/56
(B) 3/8
(C) 15/28
(D) 5/8
(E) 3/4

BrentGMATPrepNow · Accepted Answer

Set S = {2, 3, 5, 7, 11, 13, 17, 19}

Important: Notice that 7 of the values are ODD, and 1 value is EVEN.
Also recognize that the sum of any 3 odd integers will always be ODD. 
Conversely, the sum of 2 odd integers and 1 even integer will always be EVEN.   
 
So, P(sum is odd) = P(all 3 selected numbers are odd)
= P(1st selected number is odd   AND   2nd selected number is odd   AND   3rd selected number is odd)
= P(1st selected number is odd)   x   P(2nd selected number is odd)   x   P(3rd selected number is odd)
= 7/8   x   6/7   x   5/6
= 5/8

Answer: D

Aside: P(1st selected number is odd) = 7/8, since 7 of the 8 numbers in the set are odd. 
P(2nd selected number is odd) = 6/7, because once we select an odd number in the first draw, there are 7 numbers remaining, and 6 of them are odd. 
etc....

GmatPoint · Answer

The prime numbers between 0 and 20 are 2, 3, 5, 7, 11, 13, 17, and 19.
The only even prime number is 2.
If all the three numbers selected are odd, then the sum of the numbers will be odd. If one of the numbers is 2, the sum will be even.
Thus, we have to discard the cases where 2 have been selected.

If 2 is selected, the total ways to select the other two numbers will be 7C2 = 21

Total number of ways to select 3 numbers = 8C3 = 56

Total favourable cases = 56 - 21 = 35

Required probability = 35/56 = 5/8

Thus, the correct option is D.

naveeng15 · Answer

number of even numbers between 0 to 20
2,3,5,7,11,13,17,19
total 8

total possible cases = 8*7*6 , since once the number is taken the remaining will be 7 and then 6

rules of number 
odd + odd = even so 
even + odd + odd = even

so 3+3+3 will be odd

so favourable outcomes are 7*6*5 (since we should not select a even number i.e., 2 ) even + odd + odd will be even (2+3+3 =8) , so there should be no 2 in selection

so 7/8 X 6/ 7 X 5/ 6 = 210/336 = 5/8

Maria240895chile · Answer

prime numbers between 0-20=2,3,5,7,11,13,17,19= 8 numbers

They are asking how many sums of 3 numbers are odd, so we can start calculating the total number of sums using 3C8=56
in order for a sum of three numbers to be odd, we need to select 3 odd numbers ( exclude 2) so we need to select 3 numbers out of 7 using 3C7=35

the final answer is 35 out of 56  \frac{35}{56} , simplify that by 7 to get  \frac{5}{8} .

IMO (D)

Usernamevisible · Answer

Is it not assumed that repetition is not allowed ?

vasu1104 · Answer

Bunuel , why are we not considering case with even,even, odd?

is it because repetition is not allowed?
and how can we assume here that repetition is not allowed?

thanks in advance.

Bunuel · Answer

It should have been mentioned more clearly. However, the intended meaning is that repetition is not allowed.

There is only one even prime in the set: 2. So the case even, even, odd is not possible.

vasu1104 · Answer

understood.

thanks for helping as always.

egmat · Answer

Hi Usernamevisible,

Good question, and it's the same worry Legis raised in the thread: could a number be picked more than once, which would open up ODD+EVEN+EVEN as an odd-sum case?

Here's the convention to trust. When a GMAT question says you  choose  numbers  from a set , it means you are picking distinct elements - no replacement. Each pick removes that element, so you can never grab the same number twice. That's exactly why Brent's solution used  7/8 × 6/7 × 5/6 : the denominator shrinks from  8  to  7  to  6  as numbers are taken out. If repetition were allowed, every draw would stay out of  8 .

Why this settles the EVEN+EVEN worry:  set S has only  one  even number, the prime  2 . Without repetition, there's no way to select two evens - there's only one to begin with. So ODD+EVEN+EVEN simply can't happen here, and the only odd-sum case is all three odd.

A couple of signals that repetition is off unless stated:
- The wording is "chosen from set S" (selecting members of a set), not "a number from 1-20 is picked three times."
- The listed answer choice  5/8  matches the no-repetition count exactly ( 35/56 ), confirming the intended reading.

Quick way to feel the rule:  picture choosing  2  numbers from just  {2, 3} . Without replacement your only pick is {2, 3} - you can't get {2, 2}. Repetition would be a different problem, and the GMAT tells you when that's the case (e.g., "with replacement" or rolling a die repeatedly).

So your instinct is right: repetition is  not  allowed here, and that's why the EVEN+EVEN branch is off the table.

Answer: D

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