A locker can be opened by dialing a fixed three digit code (betwe… - Vidyadip

MCQ

A locker can be opened by dialing a fixed three digit code (between $000$ and $999$). A stranger who does not know the code tries to open the locker by dialing three digits at random. The probability that the stranger succeeds at the ${k^{th}}$ trial is

A
$\frac{k}{{999}}$
✓
$\frac{k}{{1000}}$
C
$\frac{{k - 1}}{{1000}}$
D
None of these

✓

Answer

Correct option: B.

$\frac{k}{{1000}}$

b
(b) Let $A$ denote the event that the stranger succeeds at the ${k^{th}}$ trial. Then

$P(A') = \frac{{999}}{{1000}} \times \frac{{998}}{{999}} \times ..... \times \frac{{1000 - k + 1}}{{1000 - k + 2}} \times \frac{{1000 - k}}{{1000 - k + 1}}$

$ \Rightarrow $$P(A')$$ = \frac{{1000 - k}}{{1000}}$

$⇒$ $P(A) = 1 - \frac{{1000 - k}}{{1000}} = \frac{k}{{1000}}.$

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

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $ A $ is a non- singular matrix, then $ A(adj A) =$

Let $A$ and $B$ be two finite sets with m and n elements respectively. The total number of subsets of the set $A$ is $56$ more than the total number of subsets of $B$. Then the distance of the point $P(m, n)$ from the point $Q (-2,-3)$ is

If $f(x) = cos \left[ {\frac{\pi }{x}} \right] cos \left( {\frac{\pi }{2}\,\,\left( {x\,\, - \,\,1} \right)} \right)$ then $f(x)$ is continuous at :

where $[x]$ is the greatest integerr function of $x$,

One of the limit point of the coaxial system of circles containing ${x^2} + {y^2} - 6x - 6y + 4 = 0$, ${x^2} + {y^2} - 2x$ $ - 4y + 3 = 0$ is

If $\alpha \ne \beta $ but ${\alpha ^2} = 5\alpha - 3$ and ${\beta ^2} = 5\beta - 3$, then the equation whose roots are $\alpha /\beta $ and $\beta /\alpha $ is

If a function $f(x)$ defined by $f(x)=\left\{\begin{array}{ll}a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4\end{array}\right.$

be continuous for some $a, b, c \in R$ and $f ^{\prime}(0)+ f ^{\prime}(2)= e ,$ then the value of of $a$ is

Let ${\tan ^{ - 1}}\left( {\tan \frac{{5\pi }}{4}} \right) = \alpha ,{\tan ^{ - 1}}\left( { - \tan \frac{{2\pi }}{3}} \right) = \beta $ Then :-

Let $f:( - 1,1) \to B$, be a function defined by $f(x) = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}},$ then $f$ is both one- one and onto when $B$ is the interval

Let $f( x )=\left\{\begin{array}{cc}-2, & -2 \leq x \leq 0 \\ x -2, & 0< x \leq 2\end{array}\right.$ and $h ( x )= f ( x \mid)+| f ( x )|$ Then $\int_{-2}^2 h(x) d x$ is equal to :

The solution of ${\log _{\sqrt 3 }}x + {\log _{\sqrt[4]{3}}}x + {\log _{\sqrt[6]{3}}}x + ......... + {\log _{\sqrt[{16}]{3}}}x = 36$ is