Question
If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{B}|\overline{\text{A}})=$

Answer

  1. $\frac{3}{5}$
Solution:
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\text{P}(\text{B}\cap\overline{\text{A}})}{\text{P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{3}{5}$

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

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:
  1. $\text{a}=\frac{1}{2},\text{b}=-\frac{3}{2}$
  2. $\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$
  3. $\text{a}=1,\text{b}=-1$
  4. None of these.
$A B C D$ is a rhombus, whose diagonals intersect at $E$. Then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals
Let a, b, c be positive real numbers. The following system of equations in x, y and z $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1,$ $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1,$ $-\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1$ has:
  1. No solution.
  2. Unique solution.
  3. Infinitely many solutions.
  4. Finitely many solutions.
If A and B are two events such that $\text{P(A)}\neq0$ and $\text{P(B)}\neq1,$ then $\text{P}(\overline{\text{A}}|\overline{\text{B}})=$
  1. $1-\text{P}(\text{A}|\text{B})$
  2. $1-\text{P}(\overline{\text{A}}|\text{B})$
  3. $\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
  4. $=\frac{\text{P}(\overline{\text{A}})}{\text{P}(\overline{\text{B}})}$
For non-zero vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good, if:
  1. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=0$
  2. $\vec{\text{a}}.\vec{\text{b}}=0=\vec{\text{c}}.\vec{\text{a}}$
  3. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
  4. $\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
Let $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1.$ Then, for what value of $\alpha$ is f(f(x)) = x?
  1. $\sqrt{2}$
  2. $-\sqrt{2}$
  3. 1
  4. -1
Which of the following are true?
  1. Particular solution is a solution of a differential equation containing no arbitrary constants.
  2. Particular Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
  3. General solution is a solution of a differential equation containing no arbitrary constants.
  4. General Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
The points of discontinuity of the function $\text{f(x)}=\begin{cases}2\sqrt{\text{x}},&0\leq\text{x}\leq1\\4-2\text{x},&1<\text{x}<\frac{5}{2}\\2\text{x}-7,&\frac{5}{2}\leq\text{x}\leq4\end{cases}$ is (are):
  1. $\text{x}=1,\text{x}=\frac{5}{2}$
  2. $\text{x}=\frac{5}{2}$
  3. $\text{x}=1,\frac{5}{2},4$
  4. $\text{x}=0,4$
Evaluate : $\cos \left(2 \cos ^{-1}\left(\frac{2}{5}\right)\right)$
The value of $k(k<0)$ for which the function $f$ defined as $f(x)=\left\{\begin{array}{cc}\frac{1-\cos k x}{x \sin x} & , x \neq 0 \\ \frac{1}{2} & , x=0\end{array}\right.$ is continuous at $x=0$ is