M08-05

Official Solution:


Two sets are defined as follows:

\(A = \{2, 3, 4, 4, 4\}\)

\(B = \{0, 1, 2\}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?


A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)


The total # of outcomes is \(5*3=15\).

Now, as set \(B\) is smaller, then work with it:

0 can be added to 2 or to 3 to get a prime: \(1+1=2\) cases;

1 can be added to 2 or to either of three 4's to get a prime: \(1+3=4\) cases;

2 can be added only to 3 to get a prime: 1 case.

So, the total # of favorable outcomes is \(2+4+1=7\). \(P=\frac{\text{favorable}}{\text{total}}=\frac{7}{15}\).


Answer: D

How come there are 15 total possible outcomes... when there are only 3 distinct elements in set A.
I believe the possible outcomes should only be 9.
plz correct me if im wrong

Two sets are defined as follows:

\(A = \{2, 3, 4, 4, 4\}\)

\(B = \{0, 1, 2\}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?


A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)

KarishmaB can you pls explain why arent we considering cases like (0;2) and (2:0) as separate ones and (2;1) and (1;2) and other similar cases