Solution:
Let, the event of numbers divisible by 3 be A
and the event of square number be B.
Then,
S = {1, 2, 3, 4, 5, 6}
A = {3, 6}
B = {1, 4}
∴ n(S) = 6, n(A) = 2 and n(B) = 2
To find: n(A∪B)
Here,
A and B are mutually exclusive events. Then by formula
P(A∪B) = p(A) + p(B)
= \(\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}\)
= \(\frac26+\frac26\)
= \(\frac23\)
∴ The probability of getting a number divisible by 3 or a square number is 2/3.