# Current Affairs JEE Main & Advanced

Category : JEE Main & Advanced

Notations : (i) $P(A+B)\text{ or }P(A\cup B)=$ Probability of happening of A or B

= Probability of happening of the events A or B or both

= Probability of occurrence of at least one event A or B

(ii) $P(AB)$ or $P(A\cap B)=$ Probability of happening of events A and B together.

(1) When events are not mutually exclusive : If A and B are two events which are not mutually exclusive, then

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

or  $P(A+B)=P(A)+P(B)-P(AB)$

For any three events A, B, C

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)$$-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$

or $P(A+B+C)=P(A)+P(B)+P(C)-P(AB)-P(BC)$$-P(CA)+P(ABC)$

(2) When events are mutually exclusive : If A and B are mutually exclusive events, then $n(A\cap B)=0$ $\Rightarrow$ $P(A\cap B)=0$

$\therefore$$P(A\cup B)=P(A)+P(B)$.

For any three events A, B, C which are mutually exclusive,

$P(A\cap B)=P(B\cap C)=P(C\cap A)=P(A\cap B\cap C)=0$

$\therefore$$P(A\cup B\cup C)=P(A)+P(B)+P(C)$.

The probability of happening of any one of several mutually exclusive events is equal to the sum of their probabilities, i.e. if ${{A}_{1}},\,{{A}_{2}}.....{{A}_{n}}$ are mutually exclusive events, then

$P({{A}_{1}}+{{A}_{2}}+...+{{A}_{n}})=P({{A}_{1}})+P({{A}_{2}})+.....+P({{A}_{n}})$

i.e. $P(\sum{{{A}_{i}}})=\sum{P({{A}_{i}})}$.

(3) When events are independent : If A and B are independent events, then $P(A\cap B)=P(A).P(B)$

$\therefore$  $P(A\cup B)=P(A)+P(B)-P(A).P(B)$.

(4) Some other theorems

(i) Let A and B be two events associated with a random experiment, then

(a) $P(\bar{A}\cap B)=P(B)-P(A\cap B)$

(b) $P(A\cap \bar{B})=P(A)-P(A\cap B)$

If $B\subset A,$ then

(a) $P(A\cap \bar{B})=P(A)-P(B)$

(b) $P(B)\le P(A)$

Similarly if $A\subset B,$ then

(a) $(\bar{A}\cap B)=P(B)-P(A)$

(b) $P(A)\le P(B)$

• Probability of occurrence of neither A nor B is

$P(\bar{A}\cap \bar{B})=P(\overline{A\cup B})=1-P(A\cup B)$

(ii) Generalization of the addition theorem : If ${{A}_{1}},\,{{A}_{2}},.....,\,{{A}_{n}}$ are $n$ events associated with a random experiment, then $P\left( \bigcup\limits_{i=1}^{n}{{{A}_{i}}} \right)=\sum\limits_{i=1}^{n}{P({{A}_{i}})}-\sum\limits_{\begin{smallmatrix} i,\,j=1 \\\,i\ne j\end{smallmatrix}}^{n}{P({{A}_{i}}\cap {{A}_{j}})}+\sum\limits_{\begin{smallmatrix} i,\,j,\,k=1 \\\,i\ne j\ne k\end{smallmatrix}}^{n}{P({{A}_{i}}\cap {{A}_{j}}\cap {{A}_{k}})}+$$...+{{(-1)}^{n-1}}P({{A}_{1}}\cap {{A}_{2}}\cap .....\cap {{A}_{n}})$.

If all the events ${{A}_{i}}(i=1,\,2...,\,n)$ are mutually exclusive, then $P\,\,\left( \bigcup\limits_{i=1}^{n}{{{A}_{i}}} \right)=\sum\limits_{i=1}^{n}{P({{A}_{i}})}$

i.e. $P({{A}_{1}}\cup {{A}_{2}}\cup ....\cup {{A}_{n}})=P({{A}_{1}})+P({{A}_{2}})+....+P({{A}_{n}})$.

(iii) Booley’s inequality : If ${{A}_{1}},\,{{A}_{2}},\,....{{A}_{n}}$ are n events associated with a random experiment, then

(a) $P\left( \bigcap\limits_{i=1}^{n}{{{A}_{i}}} \right)\ge \sum\limits_{i=1}^{n}{P({{A}_{i}})-(n-1)}$

(b) $P\left( \bigcup\limits_{i=1}^{n}{{{A}_{i}}} \right)\le \sum\limits_{i=1}^{n}{P({{A}_{i}})}$

These results can be easily established by using the Principle of mathematical induction.

#### Trending Current Affairs

You need to login to perform this action.
You will be redirected in 3 sec