
Let A and B be independent events with P(A)=0.3 and P(B)=0.4. Find
a)\[P(A\cap B)\]
b)\[P(A\cup B)\]
c)P(A|B)
d)P(B|A)
Answer
616.2k+ views
Hint: Here we need to use the basic rules and formulas for the probability of union and intersection of events and conditional probability of two events.
Complete step-by-step answer:
We shall discuss the subparts of the question in the following points
By probability theory, the probability of intersection of two events corresponds to the situation where both events have occurred. The general formula is given by
\[P(A\cap B)=P\left( A \right)\text{ }P\left( B|A \right)\text{ }=\text{ }P\left( B \right)P\left( A|B \right)\]
Where P(B|A) corresponds to the probability of the occurrence of B when A has already occurred and P(A|B) corresponds to the probability of the occurrence of A when B has already occurred. As the events are independent, the probability of occurrence of A or B is independent of occurrence of B or A respectively. Thus, P(B|A)=P(B) and P(A|B)=P(A).
Thus,
\[P(A\cap B)=P\left( A \right)\text{ }P\left( B|A \right)\text{ }=\text{ }P\left( B \right)P\left( A|B \right)=0.3\times 0.4=0.12\]
By probability theory, the probability of union of two events corresponds to the probability of occurrence of A or B or both. The general formula is given by
\[P(A\cup B)=P\left( A \right)\text{ }+P\left( B \right)\]
Using the values given in the question
\[P(A\cup B)=P\left( A \right)+P\left( B \right)=0.3+0.4=0.7\]
By probability theory, P(A|B) corresponds to the probability of the occurrence of A when B has already occurred. As the events are independent, the probability of occurrence of A is independent of occurrence of B. Thus, A is equally probable to occur and does not depend on whether B has occurred or not. Thus,
\[P\left( A|B \right)=P\left( A \right)=0.3\]
By the theory of probability, P(B|A) corresponds to the probability of the occurrence of B when A has already occurred. As the events A and B are given as independent, B is equally probable to occur and does not depend on whether A has occurred or not. Thus,
\[P\left( B|A \right)=P\left( B \right)=0.4\]
Note: We note that the rules for union and intersection of probability of two events is not the same as in set theory as in set theory, it corresponds to the space occupied by the events whereas in probability, it corresponds to the probability of the occurrence of the events.
Complete step-by-step answer:
We shall discuss the subparts of the question in the following points
By probability theory, the probability of intersection of two events corresponds to the situation where both events have occurred. The general formula is given by
\[P(A\cap B)=P\left( A \right)\text{ }P\left( B|A \right)\text{ }=\text{ }P\left( B \right)P\left( A|B \right)\]
Where P(B|A) corresponds to the probability of the occurrence of B when A has already occurred and P(A|B) corresponds to the probability of the occurrence of A when B has already occurred. As the events are independent, the probability of occurrence of A or B is independent of occurrence of B or A respectively. Thus, P(B|A)=P(B) and P(A|B)=P(A).
Thus,
\[P(A\cap B)=P\left( A \right)\text{ }P\left( B|A \right)\text{ }=\text{ }P\left( B \right)P\left( A|B \right)=0.3\times 0.4=0.12\]
By probability theory, the probability of union of two events corresponds to the probability of occurrence of A or B or both. The general formula is given by
\[P(A\cup B)=P\left( A \right)\text{ }+P\left( B \right)\]
Using the values given in the question
\[P(A\cup B)=P\left( A \right)+P\left( B \right)=0.3+0.4=0.7\]
By probability theory, P(A|B) corresponds to the probability of the occurrence of A when B has already occurred. As the events are independent, the probability of occurrence of A is independent of occurrence of B. Thus, A is equally probable to occur and does not depend on whether B has occurred or not. Thus,
\[P\left( A|B \right)=P\left( A \right)=0.3\]
By the theory of probability, P(B|A) corresponds to the probability of the occurrence of B when A has already occurred. As the events A and B are given as independent, B is equally probable to occur and does not depend on whether A has occurred or not. Thus,
\[P\left( B|A \right)=P\left( B \right)=0.4\]
Note: We note that the rules for union and intersection of probability of two events is not the same as in set theory as in set theory, it corresponds to the space occupied by the events whereas in probability, it corresponds to the probability of the occurrence of the events.
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