Question

If A and B are arbitrary events then
1) \[P(A\cap B)\ge P(A)+P(B)\]
2) \[P(A\cap B)\le P(A)+P(B)\]
3) \[P(A\cap B)=P(A)+P(B)\]
4) None of these

Answer 1

Hint: We are given a question based on the formula of probability. We have to choose a correct answer from the options given to us. We know that, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]. Using the formula, we will find the most suitable option that corresponds to A and B being arbitrary.

Complete step by step answer:
According to the question given to us, we are given two sets of arbitrary events and we have to choose the most suitable option that follows.
We will first understand what an arbitrary event is. An arbitrary event is one where the outcomes of the experiment are those which are pre-established.
So when we say that, A and B are arbitrary events, we have the possible outcomes possible for the events A and B.
For example – Tossing a coin has 2 outcomes which can be either head or tails or rolling a die which has 6 outcomes which are numbers from 1 to 6.
The probability of the events A and B occurring together can be expressed using the formula,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
We know that, probability of an event is never zero, that is always greater than zero.
We have,
\[P(A)+P(B)-P(A\cap B)\ge 0\]
We can rearrange it as,
\[\Rightarrow P(A)+P(B)\ge P(A\cap B)\]
Or we can write it as,
\[\Rightarrow P(A\cap B)\le P(A)+P(B)\]
So, the option 1, 3 and 4 are removed and we are left option 2.

So, the correct answer is “Option 2”.

Note: The other options provided resemble closely to the true value or the true expression. So, while deducing the correct option from the options provided, the inequality symbols should be carefully written and understood.