WAEC - Mathematics (2014 - No. 31)
If a number is selected at random from each of the sets p = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime
\(\frac{5}{9}\)
1\(\frac{4}{9}\)
\(\frac{1}{3}\)
\(\frac{2}{9}\)
Explanation
p = { 1, 2, 3}
Q = {2, 3, 5}
prob(prime number) = prob(1 and 2) or
= prob(2 and 3) or
= prob(3 and 2)
There are three possibilities of the sum being prime
Total possibility = 9
probability(sum being prime) = \(\frac{3}{9}\)
= \(\frac{1}{3}\)
Q = {2, 3, 5}
prob(prime number) = prob(1 and 2) or
= prob(2 and 3) or
= prob(3 and 2)
There are three possibilities of the sum being prime
Total possibility = 9
probability(sum being prime) = \(\frac{3}{9}\)
= \(\frac{1}{3}\)
Comments (0)
