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A die is thrown. Let A be the event that the number obtained is greater than 3 and B be the event that the number obtained is less than 6. Then , $P (A \cup B)$ is

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$\frac{2}{5}$

$\frac{3}{5}$

answer is D.

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Detailed Solution

since,

$\begin{array}{l} A = {4, 5, 6} and B = {1, 2, 3, 4} \\ ∴ A \cap B = {4} \\ ∴ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ = \frac{3}{6} + \frac{4}{6} - \frac{1}{6} = 1 \end{array}$

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