I think that the answer would be B
E and F are two events and that P(E)=0.3 and P(F|E)=0.5. Thus, P(E and F)=0.15
Bayes' theorem is transforming preceding probabilities into succeeding probabilities. It is based on the principle of conditional probability. Conditional probability is the possibility that an event will occur because it is dependent on another event.
P(F|E)=P(E and F)÷P(E)
It is given that P(E)=0.3,P(F|E)=0.5
Using Bayes' formula,
P(F|E)=P(E and F)÷P(E)
Rearranging the formula,
⇒P(E and F)=P(F|E)×P(E)
Substituting the given values in the formula, we get
⇒P(E and F)=0.5×0.3
⇒P(E and F)=0.15
∴The correct answer is 0.15.
If, E and F are two events and that P(E)=0.3 and P(F|E)=0.5. Thus, P(E and F)=0.15.
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Answer:
0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Mean of 0.6 times a day
7 day week, so 
What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.
In which
So
0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.
-2........................
