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

If $\Delta = \left| {\,\begin{array}{*{20}{c}}x&y&z\\p&q&r\\a&b&c\end{array}\,} \right|,$ then $\left| {\,\begin{array}{*{20}{c}}x&{2y}&z\\{2p}&{4q}&{2r}\\a&{2b}&c\end{array}\,} \right|$equals
If $a < \frac{1}{{32}},$ then the number of solution of ${({\sin ^{ - 1}}x)^3} + {({\cos ^{ - 1}}x)^3} = a{\pi ^3}$ is
The order and degree of the differential equation ${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}}$ are
If $\text{A}_{\text{r}}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix},$ then the value of $\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}$ is:
  1. n
  2. 2n
  3. -2n3
  4. n2
Area of the region bounded by rays |x| + y = 1 and X - axis is ___________.
  1. $\frac{1}{2}$
  2. $2$
  3. $1$
  4. $\frac{1}{4}$
If $y = \sin px$ and ${y_n}$ is the $n^{th}$ derivative of $y$, then $\left| {\begin{array}{*{20}{c}}
y&{{y_1}}&{{y_2}}\\
{{y_3}}&{{y_4}}&{{y_5}}\\
{{y_6}}&{{y_7}}&{{y_7}}
\end{array}} \right|$ is equal to
Let $f :[2,4] \rightarrow R$ be a differentiable function such that $\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1$, $x \in[2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements:

$(A): f(x) \leq 1$, for all $x \in[2,4]$

$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,

If ${l_1},\,{m_1},\,{n_1}$ and ${l_2},{m_2},{n_2}$ are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be
$\int_{}^{} {\frac{1}{{{{\cos }^{ - 1}}x.\sqrt {1 - {x^2}} }}dx = } $
The area bounded by the curve $y = 4x - {x^2}$ and the $x - $ axis, is