Question

A dice is rolled once. What is the probability of getting : <br />a) a 3 ?<br />b) an even number ?​

Answer 1

$\huge\underline\bold {Answer:}$

a) Number of 3’s on a dice = 1

Total number of possible outcomes = 6

(Since Sample space = {1, 2, 3, 4, 5, 6})

Therefore probability of getting 3 = 1/6.

b) Number of even numbers on a dice

= 3(2, 4, 6)

Total number of possible outcomes = 6

Therefore probability of getting an even number

= 3/6 = 1/2.

Answer 2

It’s the probability of the event “even” or “multiple of 3”, and it’s given by the sum of the probability of having respectively an even number and a number that is a multiple of 3, minus the probability that these two events occur at the same time (and this is the case if you obtain a “6”).

Formally:

P(“even” OR "multiple of 3″) = P(“even”) + P(“multiple of 3”) - P(“even” AND “multiple of 3”) = 3/6 + 2/6 - 1/6 = 4/6 = 2/3

You can verify this by counting the events in favour of the event (“even” OR “multiple of 3”) over the possible ones: the “favourable” numbers are, in our case, 2, 3, 4 and 6 (they all are or even or multiple of 3), and the possible events are all the possible outcomes, i.e. 1, 2, 3, 4, 5, and 6. So the “favourable” events are 4 and the possible events 6, so the probability of our event (“even” or “multiple of 3”) is 4 over 6, that is what we found above.