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 Suppose that A, B, C are events such that \[P(A) = P(B) = P(C) = \dfrac{1}{4}\], \[P(AB) = 0,P(AC) = \dfrac{1}{8}\], then \[P(A + B)\] equals to:
A. \[0.125\]
B. \[0.25\]
C. \[0.375\]
D. \[0.5\]


Answer
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163.5k+ views
Hint:
To solve the given question, we will use the formula of probability of the sum of two events \[P(A + B) = P(A) + P(B) - P(AB)\] and then substitute the given values in the formula to get a solution.



Formula Used:
\[P(A + B) = P(A) + P(B) - P(AB)\]


Complete step-by-step answer:
Given: \[P(A) = P(B) = P(C) = \dfrac{1}{2}\].
and \[P(AB) = 0\].
We will apply the formula of \[P(A + B) = P(A) + P(B) - P(AB)\] and substitute the given values
Therefore,
\[P(A + B) = \dfrac{1}{4} + \dfrac{1}{4} - 0\]
\[ \Rightarrow P(A + B) = \dfrac{2}{4}\]
\[ \Rightarrow P(A + B) = \dfrac{1}{2}\]
\[ \Rightarrow P(A + B) = 0.5\]

Hence, option D is correct.

Additional information:
Mutually independent: If the probability of two events does not depend on the probability of the other event, then the events are independent. Here, the probability of the intersection of two events is zero.
Mutually dependent: If the probability of two events depends on the probability of other events, then the events are dependent. Here, the probability of intersection of the event must be not equal to zero.
The probability of an event varies from 0 to 1.



Note:
Students are often confused between the sum of two events.
If the events are mutually independent then \[P(A + B) = P(A) + P(B)\].
If the event is mutually dependent then \[P(A + B) = P(A) + P(B) - P(AB)\].