MCQ
A man takes a step forward with probability $0.4$ and backward with probability $0.6.$ The probability that at the end of eleven steps he is one step away from the starting point is
  • $^{11}{C_6}\,{(0.24)^5}$
  • B
    $^{11}{C_6}{(0.4)^6}{(0.6)^5}$
  • C
    $^{11}{C_6}{(0.6)^6}\,{(0.4)^5}$
  • D
    None of these

Answer

Correct option: A.
$^{11}{C_6}\,{(0.24)^5}$
a
(a) The man will be one step away from the starting point if $(i)$ either he is one step ahead or $(ii)$ one step behind the starting point.
$\therefore $The required probability $ = P(i) + P(ii)$
The man will be one step ahead at the end of eleven steps if he moves six steps forward and five steps backward. The probability of this event $ = \,{}^{11}{C_6}{(0.4)^6}{(0.6)^5}$.
The man will be one step behind at the end of eleven steps if he moves six steps backward and five steps forward.
The probability of this event $ = \,{}^{11}{C_6}{(0.6)^6}{(0.4)^5}$.
Hence the required probability
$ = \,{}^{11}{C_6}{(0.4)^6}{(0.6)^5} + {}^{11}{C_6}{(0.6)^6}{(0.4)^5}$
$ = \,{}^{11}{C_6}{(0.4)^5}{(0.6)^5}(0.4 + 0.6) = \,{}^{11}{C_6}{(0.24)^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 $A=\left(\begin{array}{ll}{2} & {2} \\ {9} & {4}\end{array}\right)$ and $I=\left(\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right),$ then $10 A^{-1}$ is equal to
A stone is falling freely and describes a distance s in $t$ seconds given by equation $s = {1 \over 2}g\,{t^2}$. The acceleration of the stone is
The corner points of the feasible region determined by the system of linear constraints are $(0,0),(0,40),(20,40),(60,20),(60,0) .$ The objective function is $z=4 x+3 y$ Compare the quantity in Column $A$ and Column $B$
Column Maximum of $z$
$A$ $300$
$B$ $325$
Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty]$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.

(There are two questions based on $PARAGRAPH "II"$, the question given below is one of them)

($1$) The value of $2 \int^{\frac{\pi}{2}} f(x) g(x) d x-\int^{\frac{\pi}{2}} g(x) d x$ us

($2$) The value of $\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x) g(x) d x$ is

Give the answer or quetion ($1$) and ($2$) 

The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the $x$-axis and vertices $C$ and $D$ lie on the parabola, $y = x ^{2}-1$ below the $x$ -axis, is
If f : R → R is given by f(x) = x3 + 3, then f-1(x) is equal to:
  1. $\text{x}^\frac{1}{3}-3$
  2. $\text{x}^\frac{1}{3}+3$
  3. $(\text{x}-3)^\frac{1}{3}$
  4. $\text{x}+3^\frac{1}{3}$
If $\vec{a}+\vec{b}=\hat{i}$ and $\vec{a}=2 \hat{i}-2 \hat{j}+2 \hat{k}$, then $|\vec{b}|$ equals:
If the domain of the function $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$ is $[\alpha, \beta) \cup(\gamma, \delta]$, then $|3 \alpha+10(\beta+\gamma)+21 \delta|$ is equal to $.......$.
The solution of the differential equation,
$\text{x}^2\frac{\text{dy}}{\text{dx}}.\cos\frac{1}{\text{x}}-\text{y}\sin\frac{1}{\text{x}}=-1,$ where $\text{y}\rightarrow-1$ as $\text{x}\rightarrow-\infty,$ is:
  1. $\text{y}=\sin\frac{1}{\text{x}}-\cos\frac{1}{\text{x}}$
  2. $\text{y}=\frac{\text{x}+1}{\text{x}\sin\frac{1}{\text{x}}}$
  3. $\text{y}=\cos\frac{1}{\text{x}}+\sin\frac{1}{\text{x}}$
  4. $\text{y}=\frac{\text{x}+1}{\text{x}\cos\frac{1}{\text{x}}}$
lf $\text{AB}\perp\text{BC}$ then the value of $\lambda$ equal, where A(2k, 2, 3), B(k, 1, 5), C(3 + k, 2, 1):