A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, (A ∪ B) is

1

We have,

$P(A)=\frac{3}{6}=\frac{1}{2}, P(B) =\frac{4}{6}=\frac{2}{3}$

and, P(A ∩ B) = Probability of getting a number 3 and less than 5

= Probability of getting 4 = $\frac{1}{6}$

$∴ P(A ∪ B) = P(A) +P(B) - P(A ∩ B) =\frac{1}{2} +\frac{2}{3} -\frac{1}{6}= 1$