In an experiment, the probability of the event E is known to be .4. Also the probability of the event F is .8,and the probability of E∪F is 1.
i) Compute the probability of these events: P(E∩F)=
P(E-F)=
ii) Suppose the experiment is run twice in succession. What is the probability of the event E occuring at least once? Answer:?
.2
.2
.64
To compute the probability of these events, we can use the following formulas:
i) P(E∩F) is the probability of both events E and F occurring, which can be calculated using the formula P(E∩F) = P(E) + P(F) - P(E∪F). Given that P(E) = 0.4, P(F) = 0.8, and P(E∪F) = 1, we can substitute these values into the formula: P(E∩F) = 0.4 + 0.8 - 1 = 0.2.
ii) To calculate the probability of event E occurring at least once in two successive experiments, we can use the complement rule. The complement event of "E occurring at least once" is "E not occurring in both experiments." The probability of E not occurring in one experiment is 1 - P(E). As the experiment is run twice, we can multiply this probability by itself: (1 - P(E)) * (1 - P(E)). Substituting P(E) = 0.4, we get (1 - 0.4) * (1 - 0.4) = 0.6 * 0.6 = 0.36.
Therefore, the probability of event E occurring at least once in two successive experiments is 0.36.