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(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

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

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(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 = $\frac{105+210}{1024} = \frac{315}{1024} = \frac{315}{2^{10}}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

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