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