MCQ
Two numbers are chosen from {1, 2, 3, 4, 5, 6} one after another without replacement. Find the probability that the smaller of the two is less than 4.
  • $\frac{4}{5}$
  • B
    $\frac{1}{15}$
  • C
    $\frac{1}{5}$
  • D
    $\frac{14}{15}$

Answer

Correct option: A.
$\frac{4}{5}$
Total number of ways of choosing two numbers out of six $=\ ^6\text{C}_6=\frac{(6\times2)}{2}=3\times5=15$
If smaller number is chosen as 3 then greater has choice are 4, 5, 6 So, total choices = 3
If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6 So, total choices = 4
If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6 So, total choices = 5
Total favourable case = 3 + 4 + 5 = 12
Now, required probability
$=\frac{12}{15}=\frac{4}{5}$

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

The greatest value of the term independent of $x$ in the expansion of ${\left( {x\sin \theta  + \frac{{\cos \theta }}{x}} \right)^{10}}$ is
If $\left({ }^{30} C _1\right)^2+2\left({ }^{30} C _2\right)^2+3\left({ }^{30} C _3\right)^2+\ldots \ldots+30\left({ }^{30} C _{30}\right)^2=$ $\frac{\alpha 60 !}{(30 !)^2}$, then $\alpha$ is equal to
If $\text{A+B+C}=\pi,$ then $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$ is equal to:
What is the value of x if $(\text{a} + 2\text{b} – 3\text{c})\text{x}^2 + (\text{b} + 2\text{c} – 3\text{a})\text{x} + (\text{c} + 2\text{a} – 3\text{b}) = 0$ where a, b, c are in A.P?
If $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$ which is concentric with the circle $x^2+y^2+6 x+8 y-5=0$, then $\mathrm{c}=$
A circle $C$ of radius $1$ is inscribed in an equilateral triangle $P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$, respectively. The line $P Q$ is given by the equation $\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $\mathrm{C}$ are on the same side of the line $\mathrm{PQ}$.

$1.$  The equation of circle $\mathrm{C}$ is

$(A)$ $(x-2 \sqrt{3})^2+(y-1)^2=1$

$(B)$ $(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$

$(C)$ $(x-\sqrt{3})^2+(y+1)^2=1$

$(D)$ $(x-\sqrt{3})^2+(y-1)^2=1$

$2.$  Points $E$ and $F$ are given by

$(A)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$

$(B)$ $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)$

$(C)$ $\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

$(D)$ $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

$3.$  Equation of the sides $Q R, R P$ are

$(A)$ $y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$

$(B)$ $y=\frac{1}{\sqrt{3}} x, y=0$

$(C)$ $y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$

$(D)$ $y=\sqrt{3} x, y=0$

Give the answer question $1,2$ and $3.$

If $\lim _{x \rightarrow 0}\left[1+x \ln \left(1+b^2\right)\right]^{\frac{1}{x}}=2 b \sin ^2 \theta, b>0 \text { and } \theta \in(-\pi, \pi],$ then the value of $\theta$ is
The number of real solutions of the equation $|{x^2} + 4x + 3| + 2x + 5 = 0 $are
A man and a woman appear in an interview for two vacancies in the same post. The probability of man's selection is $1/4$ and that of the woman's selection is $1/3$. What is the probability that none of them will be selected
An ellipse passes through the point $(-3, 1)$ and its eccentricity is $\sqrt {\frac{2}{5}} $. The equation of the ellipse is