Question

If $ A $ and $ B $ are any two events, then the probability that exactly one of them occurs is:
(A) $ P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) $
(B) $ P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) $
(C) $ P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right) $
(D) $ P\left( A \right) + P\left( B \right) - 2P\left( {A \cup B} \right) $

Answer 1

Hint: Here, $ P(x) $ denotes the probability of some event. Thus, $ P(A \cup B) $ means the probability of $ A \cup B $ , that is any one of the events A and B occur. Similarly, $ P(A \cap B) $ means probability of $ A \cap B $ , that is both the event A and B occur. So, we are to find the probability of occurrence of exactly one of the two events A and B.
So, we will first find the probability of occurrence of any one of the two events, $ P(A \cup B) $ . Then, subtract the probability of both the events occurring together.

Complete step-by-step answer:
In the given question, we have to find the probability that exactly one of the two events A and B occur.
So, we will first find the probability that any of the two events A and B occur. So, we find the value of $ P(A \cup B) $ as we know that $ (A \cup B) $ is the set that contains all the elements of sets A and B.
Now, we know the formula $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $ .
So, we have the value of $ P(A \cup B) $ as $ P(A) + P(B) - P(A \cap B) $ .
Now, to find the value of probability such that exactly one of the events occurs, we subtract the probability of both the events occurring from the probability of any one of the events occurring.
We also know that $ (A \cap B) $ means the set of elements that belong to both the sets A and B.
So, we get, $ P\left( {exactly\,one\,of\,A\,and\,B} \right) = P(A \cup B) - P(A \cap B) $ .
Now, we know the value of $ P(A \cup B) $ as $ P(A) + P(B) - P(A \cap B) $ .
So, substituting this, we get,
 $ \Rightarrow P\left( {exactly\,one\,of\,A\,and\,B} \right) = P(A) + P(B) - P(A \cap B) - P(A \cap B) $
Now, adding up the like terms, we get,
 $ \Rightarrow P\left( {exactly\,one\,of\,A\,and\,B} \right) = P(A) + P(B) - 2P(A \cap B) $ .
So, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: These problems are the combinations of sets and probability, so, the concepts of both of the topics are used in these. Here the formula, $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $ is used. This formula is a restructured version of the formula of sets, which is, $ n(A \cup B) = n(A) + n(B) - n(A \cap B) $ where, $ n(x) $ denotes the number of elements in set $ x. $ This formula is modified into the formula of probability by dividing on both sides by $ n(U) $ , where, $ U $ is the universal set. On dividing each term by $ n(U) $ , it takes the form, $ \dfrac{{n(A \cup B)}}{{n(U)}} = \dfrac{{n(A)}}{{n(U)}} + \dfrac{{n(B)}}{{n(U)}} - \dfrac{{n(A \cap B)}}{{n(U)}} $ . The same concept is used above, and the following formula is derived, $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $ .