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Question: Let E1, E2 be two mutually exclusive events of an experiment with\[P(\bar{E}2)=0.6=P(E1\cup E2)\], t...

Let E1, E2 be two mutually exclusive events of an experiment withP(Eˉ2)=0.6=P(E1E2)P(\bar{E}2)=0.6=P(E1\cup E2), then P(E1)P(E1)
(1) 0.10.1
(2) 0.30.3
(3) 0.40.4
(4)0.20.2

Explanation

Solution

We Can solve this equation by using the suitable and easy formula of probability. Firstly find the probability of the first event and then using the suitable formula we would be able to find the probability of the second event. The outcome that has a higher probability will have higher chances to occur and the other one will be less likely to occur.

Complete step-by-step solution:
If you have proper knowledge of probability, if you know about different types of events then it would be very simple for you to solve this question.
Probability means the extent to which an event is likely to occur or how likely it is that a proposition is true.
The probability of occurring of a single event is given by P(A)P(A).
The probability of both the events to occur is given by P(A and B)P(A\text{ }and\text{ }B).
The probability of occurring one and another event is given by P(A or B)P(A\text{ }or\text{ }B)
The symbol \cup (union) means “or” that is P(E1E2)P(E1\cup E2)and it is the probability of happening of the event E1 or E2 whereas the symbol \cap (intersection) means “and” that is P(E1E2)P(E1\cap E2) is the probability of happening of the event E1 and E2.
P(E1)+P(Eˉ1)=1P(E1)+P(\bar{E}1)=1
P(E1)P(E1) is the probability of an event happening and the probability that the event does not happen will be P(Eˉ1)P(\bar{E}1).
Here you should know about mutually exclusive events to solve this question two events are said to be mutually exclusive if they cannot happen simultaneously. It is used to describe a situation where the occurrence of one outcome supersedes the other.
So, according to the question we are given that these are mutually exclusive events,
Therefore, P(E1E2)=0P(E1\cap E2)=0
We are also provided that P(Eˉ2)=0.6P(\bar{E}2)=0.6
So from the given information, we can find,
P(E2)=1P(Eˉ2)P(E2)=1-P(\bar{E}2)
That would be,
P(E2)=10.6P(E2)=1-0.6
That gives us,
P(E2)=0.4P(E2)=0.4
Now, it is given that,
P(E1E2)=0.6P(E1\cup E2)=0.6
So we have a formula,
P(E1E2)=P(E1)+P(E2)P(E1E2)P(E1\cup E2)=P(E1)+P(E2)-P(E1\cap E2)
So putting the values,
0.6=P(E1)+0.400.6=P(E1)+0.4-0
On solving the equation given above we will get ,P(E1)P(E1)
So the answer to this question is (4) 0.2.

Note: The probability of an event is a number between 0 and 1, where 0 indicates the impossibility of the event and 1 indicates certainty. If two probabilities are given then the sum of the probabilities of all possible outcomes of an experiment is 1. If the outcome comes out to be zero then the probability gives an impossible event.