Question

In a single throw of a fair die, what is the probability that an odd number or a perfect square greater than one shows up?

a. 0.6667
b. 0.5556
c. 0.75
d. 1

Answer 1

Die can show 6 numbers - from 1 to 6. Among them there are 3 odd numbers (1, 3 and 5 ) and one perfect square greater than 1 (number 4). So the amount of events that satisfy the given condition is 4 (numbers 1, 3, 4, 5).

Then the probability is

$\displaystyle P(A) = \frac{4}{6} = \frac{2}{3} = 0.6667$

Answer: a. 0.6667