MCQ
Choose the correct answer from the given four options.If $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{A}\cup\text{B})\ '+\text{P}(\text{A}\ '\cup\text{B})=$
  • A
    $\frac{1}{5}$
  • B
    $\frac{4}{5}$
  • C
    $\frac{1}{2}$
  • $1.$

Answer

Correct option: D.
$1.$
We have, $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}(\text{B})$
$=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$
Now, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A})=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2}$
$\therefore\text{P}(\text{A}\cup\text{B})\ '=1-\text{P}(\text{A}\cup\text{B})$
$=1-\frac{4}{5}=\frac{1}{5}$
And $\text{P}(\text{A}\ '\cap\text{B})=1-\text{P}(\text{A}-\text{B})$
$=1-\big[\text{P}(\text{A})-\text{P}(\text{A}\cap\text{B})\big]$
$=1-\Big(\frac{1}{2}-\frac{3}{10}\Big)=\frac{4}{5}$
$\Rightarrow\text{P}(\text{A}\cup\text{B})\ '+\text{P}(\text{A}\ '\cup\text{B})$
$=\frac{1}{5}+\frac{4}{5}=1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The point on the curve $y^2 = 4x$ which is nearest to, the point $(2, 1)$ is:
Let $A=\{1,2,3,4\}$ and $R$ be a relation in $A$ given by $R=\{(1,1),(2,2),(3,3),(4,4),(1,2)$, $(2,1),(3,1)\}$. Then, $R$ is
Let $f : N \rightarrow R$ be a function such that $f(x+y)=2 f(x) f(y)$ for natural numbers $x$ and $y$. If $f(1)=2$, then the value of $\alpha$ for which

$\sum \limits_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)$ holds, is

Let $[.]$ , $ \{.\} $ and $sgn$$(.)$ denotes greatest integer function, fractional part function and signum function respectively, then value of determinant

$\left| {\begin{array}{*{20}{c}}
  {\left[ \pi  \right]}&{amp(1 + i\sqrt 3 )}&1 \\ 
  1&0&2 \\ 
  {\operatorname{sgn} ({{\cot }^{ - 1}}x)}&1&{\{ \pi \} } 
\end{array}} \right|$ is-

The value of $\left| {\,\begin{array}{*{20}{c}}{{1^2}}&{{2^2}}&{{3^2}}\\{{2^2}}&{{3^2}}&{{4^2}}\\{{3^2}}&{{4^2}}&{{5^2}}\end{array}\,} \right|$ is
The distance of the line $\vec{\text{r}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}+\lambda(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})$ from the plane $\vec{\text{r}}.(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}})=5$ is:
Let $\text{f(x)}=\begin{cases}\frac{1}{|\text{x}|} & \text{for |x|}\geq1\\\text{ax}^2+\text{b} & \text{for |x|}<1\end{cases}$ if f(x) is continuous and differentiable at any point, then:
If $|a|\,\, = 3,\,\,\,|b|\,\, = 4$ and $|a + b|\,\, = 5,$ then $|a - b|\,\, = $
If $\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$ then, $\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)=$
The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vector is: