A man takes a step forward with probability 0 | vedclass

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Question

(A) Let $n = 11$ be the total number of steps. Let $F$ be the number of forward steps and $B$ be the number of backward steps. We have $F + B = 11$.To

Vedclass · Accepted Answer

$^{11}C_6 (0.24)^5$

Vedclass · Answer

(A) Let $n = 11$ be the total number of steps. Let $F$ be the number of forward steps and $B$ be the number of backward steps. We have $F + B = 11$.To be one step away from the starting point,the net displacement must be $F - B = 1$ or $F - B = -1$.Case $1$: $F - B = 1$. Adding $F + B = 11$ and $F - B = 1$ gives $2F = 12$,so $F = 6$ and $B = 5$.The probability is $P(F=6) = ^{11}C_6 (0.4)^6 (0.6)^5$.Case $2$: $F - B = -1$. Adding $F + B = 11$ and $F - B = -1$ gives $2F = 10$,so $F = 5$ and $B = 6$.The probability is $P(F=5) = ^{11}C_5 (0.4)^5 (0.6)^6$.Since $^{11}C_6 = ^{11}C_5$,the total probability is $P = ^{11}C_6 (0.4)^6 (0.6)^5 + ^{11}C_6 (0.4)^5 (0.6)^6$.$P = ^{11}C_6 (0.4)^5 (0.6)^5 (0.4 + 0.6) = ^{11}C_6 (0.24)^5 (1) = ^{11}C_6 (0.24)^5$.

$A$ 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

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