A die is thrown once. 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 is:

Question

A die is thrown once. 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 $P (A \cup B)$ is:

Smart Achievers · Accepted Answer

(4)

Ans. 1

Explanation:

$S = {1, 2, 3, 4, 5, 6}$

$A = {4, 5, 6}$

$B = {1, 2, 3, 4}$

$P (A) = \frac{3}{6}$

$P (B) = \frac{4}{6}$

$A \cap B = {4}$

and $P (A \cap B) = \frac{1}{6}$

We know,

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$= \frac{3}{6} + \frac{4}{6} - \frac{1}{6} = \frac{6}{6} = 1$

Solution

Q.
A die is thrown once. 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 $P (A \cup B)$ is:

1 $\frac{2}{5}$

2 $\frac{3}{5}$

3 $0$

4 $1$

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Solution

Q. A die is thrown once. 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 P(A∪B) is:

1 25

2 35

3 0

4 1