If a drunkard person tries to take a step,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $F$ be the event of a forward step,$B$ be the event of a backward step,and $S$ be the event of staying in the same position. The probabilities

Vedclass · Accepted Answer

$\frac{315}{2^{10}}$

Vedclass · Answer

(B) Let $F$ be the event of a forward step,$B$ be the event of a backward step,and $S$ be the event of staying in the same position. The probabilities are $P(F) = \frac{1}{4}$,$P(B) = \frac{1}{2}$,and $P(S) = 1 - (\frac{1}{4} + \frac{1}{2}) = \frac{1}{4}$.We want the probability that after $5$ steps,the net displacement is $1$ or $-1$.Let $n_F, n_B, n_S$ be the number of forward,backward,and stay steps respectively,where $n_F + n_B + n_S = 5$.The net displacement is $n_F - n_B = 1$ or $n_F - n_B = -1$.Case $1$: $n_F - n_B = 1$. Possible $(n_F, n_B, n_S)$ are $(1,0,4), (2,1,2), (3,2,0)$.Probabilities: $\frac{5!}{1!0!4!} (\frac{1}{4})^1 (\frac{1}{2})^0 (\frac{1}{4})^4 + \frac{5!}{2!1!2!} (\frac{1}{4})^2 (\frac{1}{2})^1 (\frac{1}{4})^2 + \frac{5!}{3!2!0!} (\frac{1}{4})^3 (\frac{1}{2})^2 (\frac{1}{4})^0 = \frac{5}{1024} + \frac{60}{1024} + \frac{40}{1024} = \frac{105}{1024}$.Case $2$: $n_F - n_B = -1$. Possible $(n_F, n_B, n_S)$ are $(0,1,4), (1,2,2), (2,3,0)$.Probabilities: $\frac{5!}{0!1!4!} (\frac{1}{4})^0 (\frac{1}{2})^1 (\frac{1}{4})^4 + \frac{5!}{1!2!2!} (\frac{1}{4})^1 (\frac{1}{2})^2 (\frac{1}{4})^2 + \frac{5!}{2!3!0!} (\frac{1}{4})^2 (\frac{1}{2})^3 (\frac{1}{4})^0 = \frac{10}{1024} + \frac{120}{1024} + \frac{80}{1024} = \frac{210}{1024}$.Total probability = $\frac{105+210}{1024} = \frac{315}{1024} = \frac{315}{2^{10}}$.

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

Similar Questions

The probability distribution of a random variable $X$ is given by the following table:
$X = x_i$ $3$ $5$ $7$ $9$
$P(X = x_i)$ $k$ $2k$ $3k$ $4k$

Then the standard deviation of $X$ is

Bismuth has a half-life of $5$ days. If a sample originally has a mass of $800 \text{ mg}$, then the mass remaining after $30$ days will be: (in $\text{ mg}$)

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to:

The random variable $X$ has a probability distribution $P(X)$ of the following form,where $k$ is some number:
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Find $P(X < 2)$,$P(X \leq 2)$,and $P(X \geq 2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module