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Conditional Probability

Category : JEE Main & Advanced

Let A and B be two events associated with a random experiment. Then, the probability of occurrence of A under the condition that B has already occurred and $P(B)\ne 0,$ is called the conditional probability and it is denoted by $P(A/B)$.

Thus, $P(A/B)=$ Probability of occurrence of A, given that B has already happened.

$=\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)}$.

Similarly, $P(B/A)=$ Probability of occurrence of B, given that A has already happened.

$=\frac{P(A\cap B)}{P(A)}=\frac{n(A\cap B)}{n(A)}$.

• Sometimes, $P(A/B)$ is also used to denote the probability of occurrence of A when B Similarly, $P(B/A)$ is used to denote the probability of occurrence of B when A occurs.

(1) Multiplication theorems on probability

(i) If A and B are two events associated with a random experiment, then$P(A\cap B)=P(A)\,.\,P(B/A)$, if $P(A)\ne 0$ or $P(A\cap B)=P(B)\,.\,P(A/B)$, if $P(B)\ne 0$.

(ii) Extension of multiplication theorem : If ${{A}_{1}},\,{{A}_{2}},\,....,\,{{A}_{n}}$ are $n$ events related to a random experiment, then $P({{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}}\cap ....\cap {{A}_{n}})=P({{A}_{1}})P({{A}_{2}}/{{A}_{1}})P({{A}_{3}}/{{A}_{1}}\cap {{A}_{2}})$$....P({{A}_{n}}/{{A}_{1}}\cap {{A}_{2}}\cap ...\cap {{A}_{n-1}})$,

where $P({{A}_{i}}/{{A}_{1}}\cap {{A}_{2}}\cap ...\cap {{A}_{i-1}})$ represents the conditional probability of the event ${{A}_{i}}$, given that the events ${{A}_{1}},\,{{A}_{2}},\,.....,\,{{A}_{i-1}}$ have already happened.

(iii) Multiplication theorems for independent events : If A and B are independent events associated with a random experiment, then $P(A\cap B)=P(A)\,.\,P(B)$ i.e., the probability of simultaneous occurrence of two independent events is equal to the product of their probabilities. By multiplication theorem, we have $P(A\cap B)=P(A)\,.\,P(B/A)$. Since A and B are independent events, therefore $P(B/A)=P(B)$. Hence, $P(A\cap B)=P(A)\,.\,P(B)$.

(iv) Extension of multiplication theorem for independent events : If ${{A}_{1}},\,{{A}_{2}},\,....,\,{{A}_{n}}$ are independent events associated with a random experiment, then

$P({{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}}\cap ...\cap {{A}_{n}})=P({{A}_{1}})P({{A}_{2}})...P({{A}_{n}})$.

By multiplication theorem, we have

$P({{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}}\cap ...\cap {{A}_{n}})=P({{A}_{1}})P({{A}_{2}}/{{A}_{1}})P({{A}_{3}}/{{A}_{1}}\cap {{A}_{2}})$$...P({{A}_{n}}/{{A}_{1}}\cap {{A}_{2}}\cap ...\cap {{A}_{n-1}})$

Since ${{A}_{1}},\,{{A}_{2}},\,....,\,{{A}_{n-1}},\,{{A}_{n}}$ are independent events, therefore

$P({{A}_{2}}/{{A}_{1}})=P({{A}_{2}}),\,P({{A}_{3}}/{{A}_{1}}\cap {{A}_{2}})=P({{A}_{3}}),\,....,$$\,P({{A}_{n}}/{{A}_{1}}\cap {{A}_{2}}\cap ...\cap {{A}_{n-1}})=P({{A}_{n}})$

Hence, $P({{A}_{1}}\cap {{A}_{2}}\cap ...\cap {{A}_{n}})=P({{A}_{1}})P({{A}_{2}})....P({{A}_{n}})$.

(2) Probability of at least one of the n independent events : If ${{p}_{1}},\,{{p}_{2}},\,{{p}_{3}},\,........,\,{{p}_{n}}$ be the probabilities of happening of n independent events ${{A}_{1}},\,{{A}_{2}},\,{{A}_{3}},\,........,\,{{A}_{n}}$ respectively, then

(i) Probability of happening none of them

$=P({{\bar{A}}_{1}}\cap {{\bar{A}}_{2}}\cap {{\bar{A}}_{3}}......\cap {{\bar{A}}_{n}})=P({{\bar{A}}_{1}}).P({{\bar{A}}_{2}}).P({{\bar{A}}_{3}}).....P({{\bar{A}}_{n}})=(1-{{p}_{1}})(1-{{p}_{2}})(1-{{p}_{3}})....(1-{{p}_{n}})$.

$=(1-{{p}_{1}})(1-{{p}_{2}})(1-{{p}_{3}})....(1-{{p}_{n}})$

(ii) Probability of happening at least one of them

$=P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}}....\cup {{A}_{n}})=1-P({{\bar{A}}_{1}})P({{\bar{A}}_{2}})P({{\bar{A}}_{3}})....P({{\bar{A}}_{n}})=1-(1-{{p}_{1}})(1-{{p}_{2}})(1-{{p}_{3}})...(1-{{p}_{n}})$.$=1-(1-{{p}_{1}})(1-{{p}_{2}})(1-{{p}_{3}})...(1-{{p}_{n}})$

(iii) Probability of happening of first event and not happening of the remaining $=P({{A}_{1}})P({{\bar{A}}_{2}})P({{\bar{A}}_{3}}).....P({{\bar{A}}_{n}})$

$={{p}_{1}}(1-{{p}_{2}})(1-{{p}_{3}}).......(1-{{p}_{n}})$

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