1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# If A and B are two events such that P $\left(\overline{)A}\cup \overline{)B}\right)=\frac{2}{3}\mathrm{and}P\left(A\cup B\right)=\frac{5}{9},\mathrm{then}P\left(\overline{)A}\right)+P\left(\overline{)B}\right)$ = ______________.

Open in App
Solution

## Given, for two events A and B $P\left(\overline{A}\cup B\right)=\frac{2}{3}\mathrm{and}P\left(A\cup B\right)=\frac{5}{9}$ $\mathrm{Since}P\left(\overline{A}\cup \overline{B}\right)=P\left(\overline{A}\right)+P\left(\overline{B}\right)-P\left(\overline{A}\cap \overline{B}\right)\phantom{\rule{0ex}{0ex}}=P\left(\overline{A}\right)+P\left(\overline{B}\right)-P\left(\left(\overline{A\cup B}\right)\right)\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}P\left(\overline{A}\cup \overline{B}\right)=P\left(\overline{A}\right)+P\left(\overline{B}\right)-\left[1-P\left(A\cup B\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}P\left(\overline{A}\right)+P\left(\overline{B}\right)=P\left(\overline{A}\cup \overline{B}\right)+1-P\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{3}+1-\frac{5}{9}\phantom{\rule{0ex}{0ex}}=\frac{6+9-5}{9}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}P\left(\overline{A}\right)+P\left(\overline{B}\right)=\frac{10}{9}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Introduction to Probability
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program