Question

If $ A $ and $ B $ are two events such that $ P(A|B) = 0.6 $ , $ P(B|A) = 0.3 $ , $ P(A) = 0.1 $ then $ P(\overline A \cap \overline B ) = $ (here $ \overline E $ is complement of the event)
a. 0.88
b. 0.12
c. 0.6
d. 0.4

Answer 1

Hint: Here the question is related to the probability. The probability is the number of selecting the terms from a given set. Here we use the conditional probability formula to find the solution. To obtain the solution the demorgan’s law is implemented for the question.

Complete step-by-step answer:
Let us consider the given data, so we have $ P(A|B) = 0.6 $ , $ P(B|A) = 0.3 $ , $ P(A) = 0.1 $ and we have to find the $ P(\overline A \cap \overline B ) $ . First, we obtain the $ P(A \cup B) $ and then $ P(\overline A \cap \overline B ) $ .
We know about the conditional probability that is $ P(B|A) = \dfrac{{P(A \cap B)}}{{P(A)}} $
Rearranging the terms, we have
By applying the data, we have $ \Rightarrow P(B|A)P(A) = P(A \cap B) $
Substituting the values
 $ \Rightarrow P(A \cap B) = 0.3 \times 0.1 $
 $ \Rightarrow P(A \cap B) = 0.03 $
And now we have to find the $ P(A \cup B) $ , to find this we have formula and that is $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $ . Firstly, we need to find the $ P(B) $ . So, let us use the conditional probability formula that is $ P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}} $
Rearranging the term, we can rewrite as
 $ \Rightarrow P(B) = \dfrac{{P(A \cap B)}}{{P(A|B)}} $
Substituting the given values
 $ \Rightarrow P(B) = \dfrac{{0.3}}{{0.6}} $
On simplification we have
 $ \Rightarrow P(B) = 0.5 $
So now we use the formula $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $
Substituting the values, we have
 $ \Rightarrow P(A \cup B) = 0.1 + 0.5 - 0.03 $
On simplification we have
 $ \Rightarrow P(A \cup B) = 0.6 - 0.03 $
 $ \Rightarrow P(A \cup B) = 0.57 $
Hence, we obtained the $ P(A \cup B) $
We know the Demorgan’s law that is $ \overline {A \cup B} = \overline A \cap \overline B $
And hence we have to find the probability we have the formula $ P\left( {\overline {A \cup B} } \right) = P\left( {\overline A \cap \overline B } \right) $ . We have found the $ P(A \cup B) $ by using this we have to use another formula and that is $ P\left( {\overline {A \cup B} } \right) = 1 - P(A \cup B) $
By substituting the value, we have
 $ \Rightarrow P\left( {\overline {A \cup B} } \right) = 1 - 0.57 $
On simplification we have
 $ \Rightarrow P\left( {\overline {A \cup B} } \right) = 0.43 $
Hence, we obtained the solution as 0.43 and we have options given in the question and we can say that the option d is the correct one.
So, the correct answer is “Option D”.

Note: Since probability is a topic which is similar to the set, we can use some property of the sets. We should know about the conditional probability and some property of the probability to find the solution. By using the formulas and property we obtained the solution.