A man takes a step forward with probability 0 | vedclass

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(B) Let $p$ be the probability of taking a step forward,so $p = 0.4$. Let $q$ be the probability of taking a step backward,so $q = 0.6$. Total steps $

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$^{11}C_{6} 	imes (0.24)^{5}$

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(B) Let $p$ be the probability of taking a step forward,so $p = 0.4$. Let $q$ be the probability of taking a step backward,so $q = 0.6$. Total steps $n = 11$. Let $x$ be the number of forward steps and $y$ be the number of backward steps. We have $x + y = 11$. For the man to be one step away from the starting point,the net displacement must be $x - y = 1$ or $x - y = -1$. Case $1$: $x - y = 1$. Adding the equations,$2x = 12 \implies x = 6$,so $y = 5$. The probability for this case is $^{11}C_{6} 	imes p^{6} 	imes q^{5} = ^{11}C_{6} 	imes (0.4)^{6} 	imes (0.6)^{5}$. Case $2$: $x - y = -1$. Adding the equations,$2x = 10 \implies x = 5$,so $y = 6$. The probability for this case is $^{11}C_{5} 	imes p^{5} 	imes q^{6} = ^{11}C_{5} 	imes (0.4)^{5} 	imes (0.6)^{6}$. Since $^{11}C_{6} = ^{11}C_{5}$,the total probability is $^{11}C_{6} 	imes (0.4)^{5} 	imes (0.6)^{5} 	imes (0.4 + 0.6) = ^{11}C_{6} 	imes (0.24)^{5} 	imes 1 = ^{11}C_{6} 	imes (0.24)^{5}$.

$A$ man takes a step forward with probability $0.4$ and one step backward with probability $0.6$. Then the probability that at the end of eleven steps he is one step away from the starting point is

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