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....

Legis · Answer

When should we assume that the numbers are taken out? The problem does not state that the numbers cannot be repeated.

If they can be repeated, would we also include the sum of (ODD+EVEN+EVEN), since that yields an odd number?

RastogiSarthak99 · Answer

If you've figured out that only odd + odd + odd = odd then the sum becomes more familiar.

total numbers in the set = 8 
total odd numbers = 7 (3,5,7...19)
total even numbers = 1 (only 2)

Fav outcomes/Total outcomes

choosing all odd/ total number of outcomes

7c3/8c3 = 5/8

the expression above basically means you choose all 3 as odd from 7 available numbers and divide it by 8c3 which is all outcomes from all available numbers

Bunuel  - I have a query, although it won't make different to the answer, is it more accurate to write (7c3 * 1c0)/8c3?

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)

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

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