VJesus12 wrote: ↑Sun Dec 26, 2021 2:54 pm
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) \(\dfrac{15}{56}\)
(B) \(\dfrac38\)
(C) \(\dfrac{15}{28}\)
(D) \(\dfrac58\)
(E) \(\dfrac34\)
Answer:
D
Source: Manhattan GMAT
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....Brent Hanneson - Creator of GMATPrepNow.com
