Search any question & find its solution
Question:
Answered & Verified by Expert
If $A$ and $B$ are events such that $P(A)=0.42, P(B)$ $=0.48$ and $P(A$ and $B)=0.16 .$ then
I. $P(\operatorname{not} A)=0.58$
II. $P(\operatorname{not} B)=0.52$
III. $\mathrm{P}(\mathrm{A}$ or $\mathrm{B})=0.47$
Options:
I. $P(\operatorname{not} A)=0.58$
II. $P(\operatorname{not} B)=0.52$
III. $\mathrm{P}(\mathrm{A}$ or $\mathrm{B})=0.47$
Solution:
1602 Upvotes
Verified Answer
The correct answer is:
Only I and II are correct.
I. $\mathrm{P}(\operatorname{not} \mathrm{A})=1-0.42=0.58$
II. $\mathrm{P}($ not $\mathrm{B})=1-\mathrm{P}(\mathrm{B})=1-0.48=0.52$
III. $\mathrm{P}(\mathrm{A}$ or $\mathrm{B})=\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.42+0.48-0.16=0.74$
II. $\mathrm{P}($ not $\mathrm{B})=1-\mathrm{P}(\mathrm{B})=1-0.48=0.52$
III. $\mathrm{P}(\mathrm{A}$ or $\mathrm{B})=\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.42+0.48-0.16=0.74$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.