To find the probability [tex]\( P(A \cup B) \)[/tex], we first need to understand the events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] given in the question:
1. Event [tex]\( A \)[/tex]: The number obtained is greater than 3.
Possible outcomes for [tex]\( A \)[/tex] when a die is thrown: 4, 5, 6.
Therefore, [tex]\( A \text{ outcomes} = \{4, 5, 6\} \)[/tex].
2. Event [tex]\( B \)[/tex]: The number obtained is less than 5.
Possible outcomes for [tex]\( B \)[/tex] when a die is thrown: 1, 2, 3, 4.
Therefore, [tex]\( B \text{ outcomes} = \{1, 2, 3, 4\} \)[/tex].
To find the probabilities of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we proceed as follows:
- The total number of outcomes when a die is rolled is 6.
- Probability of [tex]\( A \)[/tex]:
[tex]\[
P(A) = \frac{\text{Number of outcomes favorable to } A}{\text{Total number of outcomes}} = \frac{3}{6} = 0.5
\][/tex]
- Probability of [tex]\( B \)[/tex]:
[tex]\[
P(B) = \frac{\text{Number of outcomes favorable to } B}{\text{Total number of outcomes}} = \frac{4}{6} = \frac{2}{3} \approx 0.6667
\][/tex]
Next, we find the intersection of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], which is the event where the number obtained is both greater than 3 and less than 5:
- Intersection [tex]\( A \cap B \)[/tex]:
The only number that satisfies both conditions is 4.
Therefore, [tex]\( A \cap B \text{ outcomes} = \{4\} \)[/tex].
- Probability of [tex]\( A \cap B \)[/tex]:
[tex]\[
P(A \cap B) = \frac{\text{Number of outcomes favorable to } A \cap B}{\text{Total number of outcomes}} = \frac{1}{6} \approx 0.1667
\][/tex]
Finally, to find [tex]\( P(A \cup B) \)[/tex], we use the formula for the probability of the union of two events:
[tex]\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\][/tex]
Substituting the probabilities we have calculated:
[tex]\[
P(A \cup B) = 0.5 + 0.6667 - 0.1667 \approx 1
\][/tex]
Therefore, the probability [tex]\( P(A \cup B) \)[/tex] is approximately 1.
So the correct answer is:
b. 1
