Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

15.(ii) If E and F are events such that $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$ find (ii) P(not E and not F).

Answers (1)

best_answer

Given, $P(E)=\frac{1}{4}$ , $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find : $P(not\ E\ and\ not\ F) = P(E' \cap F')$

We know,

$P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)$

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

$\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$

$\implies$ $P(E' \cap F') = 1 - P(E \cup F)$

$= 1- \frac{5}{8}= \frac{3}{8}$

Therefore, $P(E' \cap F') =$ $\frac{3}{8}$

Posted by

HARSH KANKARIA

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10, 12 board exams 2025 from April 7 to 11 for sports students

anam-khan

CBSE Class 12 history paper 2025 moderately difficult, tested students’ understanding

anam-khan

CBSE Class 12 Computer Science Exam Analysis 2025: Paper 'moderate', but 'tougher' than last three years

Latest Question