Math  /  Data & Statistics

QuestionLet S={E1,E2,E3,E4}S=\left\{E_{1}, E_{2}, E_{3}, E_{4}\right\} be the sample space of an experiment. Event A={E1,E2}A=\left\{E_{1}, E_{2}\right\}. Event B={E3}B=\left\{E_{3}\right\}. Event C={E2,E3}C=\left\{E_{2}, E_{3}\right\}. The probàbilities of the sample points are assigned as follows: \begin{tabular}{cc} \hline Sample point & Probability \\ \hlineE1E_{1} & 0.1496 \\ E2E_{2} & 0.1852 \\ E3E_{3} & 0.2457 \\ E4E_{4} & 0.4195 \\ \hline \end{tabular}
Then, P(AB)P(A \cup B) is equal to
Select one: a. 0.3953 b. 0.0000 c. 0.1496 d. 0.2457 e. 0.4309 f. 0.3348 g. 0.5805 h. 1.0000 i. 0.1852

Studdy Solution
Use the formula for the probability of the union of two events to find P(AB) P(A \cup B) :
P(AB)=P(A)+P(B)P(AB) P(A \cup B) = P(A) + P(B) - P(A \cap B)
Substitute the known values:
P(AB)=0.3348+0.24570 P(A \cup B) = 0.3348 + 0.2457 - 0
P(AB)=0.5805 P(A \cup B) = 0.5805
The probability P(AB) P(A \cup B) is:
0.5805 \boxed{0.5805}

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