A man takes a step forward with probability 0 | vedclass

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(C) Let a step forward be a success $(p = 0.4)$ and a step backward be a failure $(q = 0.6)$.To be one step away from the starting point after $11$ st

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

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(C) Let a step forward be a success $(p = 0.4)$ and a step backward be a failure $(q = 0.6)$.To be one step away from the starting point after $11$ steps,the number of forward steps $(n_f)$ and backward steps $(n_b)$ must satisfy $n_f - n_b = 1$ or $n_b - n_f = 1$.Since $n_f + n_b = 11$,the possible cases are:Case $1$: $n_f = 6$ and $n_b = 5$.Case $2$: $n_f = 5$ and $n_b = 6$.The required probability is $P = { }^{11} C_6 p^6 q^5 + { }^{11} C_5 p^5 q^6$.Since ${ }^{11} C_6 = { }^{11} C_5$,we have:$P = { }^{11} C_6 (p^6 q^5 + p^5 q^6) = { }^{11} C_6 p^5 q^5 (p + q)$.Given $p + q = 0.4 + 0.6 = 1$,we get:$P = { }^{11} C_6 (0.4)^5 (0.6)^5 (1) = { }^{11} C_6 (0.24)^5$.

$A$ man takes a step forward with probability $0.4$ and backwards 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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