
If $A$ and B are two events such that $A\subseteq B$ , then $P\left( {}^{B}/{}_{A} \right)$ ?
Answer
541.5k+ views
Hint: A set $A$ is a subset of another set $B$ if all elements of the set $A$ are elements of the set $B$. So we can also describe that set $A$ is contained inside the set $B$. Since set $B$contains elements which are not present in $A$, we can say that $A$ is a proper subset of $B$. But not the other way round. And in the question, we are given that $A$ is a subset of $B$. So the probability of the intersection of the happening of $A$ and $B$ is the probability of happening of $A$.
Complete step by step solution:
Since $A$ is a subset of $B$, we can safely arrive at a conclusion with respect to their probabilities.
The conclusion is :
$\Rightarrow P\left( A\cap B \right)=P\left( A \right)$
In the question, we are asked about conditional probability.
Conditional probability is a measure of the probability of an event occurring, given that another event has already occurred.
We know the basic formula behind conditional probability. It is the following :
$\Rightarrow P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ .
Let us use this formula to solve our question.
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$
Since it is specified in the question $A$ is a subset of $B$,our conclusion was $P\left( A\cap B \right)=P\left( A \right)$.
Let us substitute this and get the answer.
Upon substituting, we get the following :
$\begin{align}
& \Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)} \\
& \Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( A \right)}{P\left( A \right)} \\
& \Rightarrow P\left( {}^{B}/{}_{A} \right)=1 \\
\end{align}$
$\therefore $ If $A$ and $B$are two events such that $A\subseteq B$ , then $P\left( {}^{B}/{}_{A} \right)=1$.
Note: It is very important to remember all the theorems in probability . We should be able to prove theorems such as the addition theorem, Bayes theorem. Problems from probability need a lot of practice. There is a lot of logic which is involved behind every problem. We should understand each and every step of the solution to be able to solve any kind of question from chapter. We should remember all the formulae and definitions as well.
Complete step by step solution:
Since $A$ is a subset of $B$, we can safely arrive at a conclusion with respect to their probabilities.
The conclusion is :
$\Rightarrow P\left( A\cap B \right)=P\left( A \right)$
In the question, we are asked about conditional probability.
Conditional probability is a measure of the probability of an event occurring, given that another event has already occurred.
We know the basic formula behind conditional probability. It is the following :
$\Rightarrow P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ .
Let us use this formula to solve our question.
$\Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$
Since it is specified in the question $A$ is a subset of $B$,our conclusion was $P\left( A\cap B \right)=P\left( A \right)$.
Let us substitute this and get the answer.
Upon substituting, we get the following :
$\begin{align}
& \Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)} \\
& \Rightarrow P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( A \right)}{P\left( A \right)} \\
& \Rightarrow P\left( {}^{B}/{}_{A} \right)=1 \\
\end{align}$
$\therefore $ If $A$ and $B$are two events such that $A\subseteq B$ , then $P\left( {}^{B}/{}_{A} \right)=1$.
Note: It is very important to remember all the theorems in probability . We should be able to prove theorems such as the addition theorem, Bayes theorem. Problems from probability need a lot of practice. There is a lot of logic which is involved behind every problem. We should understand each and every step of the solution to be able to solve any kind of question from chapter. We should remember all the formulae and definitions as well.
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