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Question: The probability that at least one of the event A and B occurs is 0.6, If A and B occur simultaneousl...

The probability that at least one of the event A and B occurs is 0.6, If A and B occur simultaneously with probability 0.2, then P(Aˉ)+P(Bˉ)P\left( {\bar A} \right) + P\left( {\bar B} \right) is
(a) 0.4
(b) 0.8
(c) 1.2
(d) 1.4

Explanation

Solution

Hint – In this question at least one event occurs means P(AB)P\left( {A \cup B} \right) and two events occur simultaneously means P(AB)P\left( {A \cap B} \right). Use the concept that probability of addition up of an event and non-occurrence of one event is always equal to one that is P(A)+P(Aˉ)=1P\left( A \right) + P\left( {\bar A} \right) = 1. This will help getting the answer.

Complete step-by-step answer:
Given data:
Probability that at least one of the events A and B occurs is 0.6.
P(AB)=0.6\Rightarrow P\left( {A \cup B} \right) = 0.6
And if A and B occur simultaneously the probability is 0.2.
P(AB)=0.2\Rightarrow P\left( {A \cap B} \right) = 0.2
So we have to find out the value of P(Aˉ)+P(Bˉ)P\left( {\bar A} \right) + P\left( {\bar B} \right), where P(Aˉ)P\left( {\bar A} \right) and P(Bˉ)P\left( {\bar B} \right) is the probability of not occurring the events A and B.
As we know that the total probability is 1 (i.e. the sum of probability of occurring and the probability of not occurring).
P(A)+P(Aˉ)=1\Rightarrow P\left( A \right) + P\left( {\bar A} \right) = 1
P(A)=1P(Aˉ)\Rightarrow P\left( A \right) = 1 - P\left( {\bar A} \right).................. (1)
Similarly,
P(B)=1P(Bˉ)\Rightarrow P\left( B \right) = 1 - P\left( {\bar B} \right)................ (2)
Now according to set relation we have,
P(AB)=P(A)+P(B)P(AB)\Rightarrow P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)
Now from equation (1) and (2) we have,
P(AB)=1P(Aˉ)+1P(Bˉ)P(AB)\Rightarrow P\left( {A \cup B} \right) = 1 - P\left( {\bar A} \right) + 1 - P\left( {\bar B} \right) - P\left( {A \cap B} \right)
P(Aˉ)+P(Bˉ)=2P(AB)P(AB)\Rightarrow P\left( {\bar A} \right) + P\left( {\bar B} \right) = 2 - P\left( {A \cup B} \right) - P\left( {A \cap B} \right)
Now substitute the values we have,
P(Aˉ)+P(Bˉ)=20.60.2=20.8=1.2\Rightarrow P\left( {\bar A} \right) + P\left( {\bar B} \right) = 2 - 0.6 - 0.2 = 2 - 0.8 = 1.2
So this is the required answer.
Hence option (C) is the correct answer.

Note – P(Aˉ)P\left( {\bar A} \right) means the probability of non-occurrence of event A. So P(Aˉ)+P(Bˉ)P\left( {\bar A} \right) + P\left( {\bar B} \right) means that we were to find the probability of non-occurrence of two events A and B. The probability is always in between 0 to 1 that means 0P(A)10 \leqslant P(A) \leqslant 1, thus the compliment or non-occurrence of any event will simply be 1P(A)1 - P(A).