Question

Let A and B be two non-null events such that $A\subset B$, then which of the following statements is always correct?
(a) $P\left( A|B \right)=1$,
(b) $P\left( A|B \right)=P\left( B \right)-P\left( A \right)$,
(c) $P\left( A|B \right)\le P\left( A \right)$,
(d) $P\left( A|B \right)\ge P\left( A \right)$.

Answer 1

Hint: We start solving the problem by writing the range for probabilities of events A and B. We then recall the definition of a subset to find the intersection of the events A and B, also the relation of A and B. We then use the definition of the conditional probability of two events and use all the obtained conditions of probabilities of events A and B to get the desired answer.

Complete step by step answer:
According to the problem, we have two non-null events A and B such that $A\subset B$. We need to check which of the given options will be true for any event A and B.
Since the given events A and B are non-null events, the probability of occurrence of them is not zero.
So, we have $0From the problem, we are given that event A is a subset of B. This means that every possibility of event A is present in event B without fail, but every possibility of event B is not in event A. So, we get the intersection of two events A and B as A and the probability of occurrence of event B is greater than or equal to the probability of occurrence of event A.
$\Rightarrow P\left( A\cap B \right)=P\left( A \right)$ ---(2).
$\Rightarrow P\left( B \right)\ge P\left( A \right)$ ---(3).
We know that the conditional probability of two events A and B is defined as $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.
From equation (2), we get $P\left( A|B \right)=\dfrac{P\left( A \right)}{P\left( B \right)}$.
From equation (3), we get $P\left( A|B \right)\le \dfrac{P\left( A \right)}{P\left( A \right)}$.
$\Rightarrow P\left( A|B \right)\le 1$ ---(4).
Now, we have $P\left( A|B \right)=\dfrac{P\left( A \right)}{P\left( B \right)}$.
$\Rightarrow P\left( B \right)=\dfrac{P\left( A \right)}{P\left( A|B \right)}$ ---(5).
From equation (1) we have $P\left( B \right)\le 1$.
$\Rightarrow \dfrac{P\left( A \right)}{P\left( A|B \right)}\le 1$.
$\Rightarrow P\left( A \right)\le P\left( A|B \right)$.
$\Rightarrow P\left( A|B \right)\ge P\left( A \right)$ ---(6).
From equations (4) and (6), we have found $P\left( A|B \right)\ge P\left( A \right)$.
∴ The correct option for the given problem is (d).

Note:
 We should know that if $x\ge 0$, then $\dfrac{1}{x}\le 0$. We used this property to find the result of $P\left( A|B \right)\le 1$. Since event A is the subset of event B, the probability of occurrence of B will sometimes be equal to the probability of occurrence of A. We can also verify this result by taking an example for the events A and B. We should make the operations carefully while doing this problem.