$\begin{array}{1 1}(A)\;0.4\\(B)\;0.8\\(C)\;1.2\\(D)\;1.6\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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

Step 1:

Given :

$P(A \cup B)=0.6,P(A \cap B)=0.2$

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

$0.6=P(A)+P(B)-0.2$

$\Rightarrow 0.6+0.2=P(A)+P(B)$

$\Rightarrow P(A)+P(B)=0.8$

Step 2:

$P(\bar{A})+P(\bar{B})=(1-P(A))+1-P(B))$

$\Rightarrow 2-(P(A)+P(B))$

$\Rightarrow 2-0.8=1.2$

Hence (C) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...