Problem Solving

swerve wrote: ↑

Mon Jan 27, 2020 6:06 pm

\(A: \{71, 73, 79, 83, 87\}\)
\(B: \{57, 59, 61, 67\}\)

If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?

A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. \(1\)

The OA is B

Source: Magoosh

Note that between 50 to 100, the prime numbers are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

• Prime numbers in Set A: 71, 73, 79, 83; there are four primes
Total numbers in Set A = 4

Thus, the probability of selecting a prime from Set A = 4/5;

• Prime numbers in Set B: 59, 61, 67; there are three primes
Total numbers in Set B = 3

Thus, the probability of selecting a prime from Set B = 3/4;

Thus, the probability that both numbers are prime = 4/5*3/4 = 3/5

The correct answer: B

Hope this helps!

-Jay
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swerve wrote: ↑

Mon Jan 27, 2020 6:06 pm

\(A: \{71, 73, 79, 83, 87\}\)
\(B: \{57, 59, 61, 67\}\)

If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?

A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. \(1\)

The OA is B

Source: Magoosh

Since there are 4 primes in set A (all are primes except 87, which is divisible by 3) and there are 3 primes in set B (all are primes except 57, which is also divisible by 3), the probability that both selected numbers are prime is 4/5 x 3/4 = 3/5.

Answer: B