MCQ
A four - digit number is formed by using the digits 1, 2, 4, 8 and 9 without repitition. If one number is selected from those numbers, then what is the probability that it will be an odd number?
  • A
    $\frac{1}{5}$
  • B
    $\frac{2}{5}$
  • C
    $\frac{3}{5}$
  • D
    $\frac{4}{5}$

Answer

  1. $\frac{2}{5}$

Solution:

Total number of outcomes = 5 × 4 × 3 × 2 = 120

he number of favourable cases = 2 (4 × 3 × 2) - = 48 (i.e., odd numbers)

herefore,Required probability $=\frac{48}{120}=\frac{2}{5}.$

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

 Area bounded by the curve $\text{y}=\sin\text{x}$ and the x-axis between $\text{x}=0$ and $\text{x}=2\pi$ is:
  1. 2 sq units
  2. 0 sq units
  3. 3 sq units
  4. 4 sq units
If $y=5 \cos x-3 \sin x$, then $\frac{d^2 y}{d x^2}$ is equal to
Let $[ t ]$ denotes the greatest integer $\leq t$. Then $\frac{2}{\pi} \int \limits_{\pi / 6}^{5 \pi / 6}(8[\operatorname{cosec} x]-5[\cot x]) d x$ is equal to
Let $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$ Let $\vec{d}$ be a vector which is perpendicular to both $\overrightarrow{ a }$ and $\overrightarrow{ b }, \quad$ and $\quad \overrightarrow{ c } \cdot \overrightarrow{ d }=12$. Then $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$ is equal to $........$.
Let $l_{1}$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4 \sqrt{2}}$ respectively, and $l_{2}$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6 \sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_{1}$ and $l_{2}$, then $d ^{-2}$ is equal to
$\int_{}^{} {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \;dx = $
$\int_{}^{} {\sqrt {\frac{x}{{{a^3} - {x^3}}}} \;dx = } $
If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
  {x + 2}&\omega &{{\omega ^2}} \\ 
  \omega &{x + 1 + {\omega ^2}}&1 \\ 
  {{\omega ^2}}&1&{x + 1 + \omega } 
\end{array}} \right| = 0$ is 
If integrating factor of $x(1 - {x^2})dy + (2{x^2}y - y - a{x^3})dx = 0$ is ${e^{\int_{}^{} {Pdx} }},$ then $P$ is equal to
Box $I$ contains $30$ cards numbered $1$ to $30$ and Box $II$ contains $20$ cards numbered $31$ to $50 .$ A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is