Number \(N\) is randomly selected from a set of all primes between \(10\) and \(40,\) inclusive. Number \(K\) is selecte

VJesus12 wrote: ↑

Sun Aug 15, 2021 8:44 am

Number \(N\) is randomly selected from a set of all primes between \(10\) and \(40,\) inclusive. Number \(K\) is selected from a set of all multiples of \(5\) between \(10\) and \(40\) inclusive. What is the probability that \(N+K\) is odd?

(A) \(\dfrac12\)

(B) \(\dfrac23\)

(C) \(\dfrac34\)

(D) \(\dfrac47\)

(E) \(\dfrac58\)

Answer: D

Source: Veritas Prep

Notice that there are 2 ways that the sum N+K can be ODD
1) N is ODD and K is EVEN
2) N is EVEN and K is ODD

However, N cannot be EVEN, since all of the primes between 10 and 40 are ODD (11, 13, 17, 19, . . . . 31, 37)
So, case 2 (above) is impossible

So, P(N+K is odd) = P(N is odd AND K is even)
= P(N is odd) x P(K is even)

------ASIDE-------
Possible values of N: (11, 13, 17, 19, . . . . 31, 37)
So, P(N is odd) = 1

Possible values of K: (10, 15, 20, 25, 30, 35, 40)
So, P(K is even) = 4/7
-----------------------------------------

So, P(N+K is odd) = P(N is odd AND K is even)
= P(N is odd) x P(K is even)
= 1 x 4/7
= 4/7
= D

Cheers,
Brent