An experiment in which the three mutually exclusive events, A, B, and C form a partition of the uniform sample space S is depicted in the diagram.

Find .

Question

An experiment in which the three mutually exclusive events, A, B, and C form a partition of the uniform sample space S is depicted in the diagram. ​
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Find  .
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Quizplus · Accepted Answer

From the given diagram, we can see that the probability of event A is 0.4, the probability of event B is 0.3, and the probability of event C is 0.2. Therefore, the probability of the complement of event A (i.e., not A) is 0.6, the probability of the complement of event B (i.e., not B) is 0.7, and the probability of the complement of event C (i.e., not C) is 0.8. The probability of the union of events A and B is 0.6 (i.e., the probability of the complement of event C), and the probability of the intersection of events A and B is 0.2. Therefore, the probability of the event (A intersection B) complement (i.e., not A or not B) is 0.8 - 0.2 = 0.6. Finally, we can use the law of total probability to find the probability of event A or event B or event C, which is given by P(A or B or C) = P(A) + P(B) + P(C) - P(A intersection B) - P(A intersection C) - P(B intersection C) + P(A intersection B intersection C) = 0.4 + 0.3 + 0.2 - 0.2 - 0 - 0 + 0 = 0.7. Therefore, the probability of the complement of event (A or B or C) is 0.3, which is the same as the probability of event (not A and not B and not C). Hence, the answer is (B) 11eab448_e7e9_b705_acdb_3576376f5098_TB6027_11.

An Experiment in Which the Three Mutually Exclusive Events, A