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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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Verified Answer
The correct answer is:
${ }^{11} \mathrm{C}_6(0.4)^6(0.6)^5$
Let a step forward be a success and the step backward be a failure.
$\therefore \quad$ Probability of success $=\mathrm{p}=0.4$, and Probability of failure $=q=0.6$
Now, in 11 steps number of successes $=6$, number of failure $=5$
OR
number of successes $=5$, number of failures $=6$
Required probability $={ }^{11} \mathrm{C}_6 \mathrm{p}^6 \mathrm{q}^5+{ }^{11} \mathrm{C}_5 \mathrm{p}^5 \mathrm{q}^6$.
$\begin{aligned}
& =\frac{11 !}{6 ! 5 !} p^6 q^5+\frac{11 !}{5 ! 6 !} p^5 q^6 \\
& ={ }^{11} C_6 p^5 q^5(p+q) \\
& ={ }^{11} C_6(0-4)^5(0-6)^5(1) \\
& ={ }^{11} C_6(0 \cdot 4)^5(0 \cdot 6)^5 \\
& ={ }^{11} C_6(0 \cdot 24)^5
\end{aligned}$
$\therefore \quad$ Probability of success $=\mathrm{p}=0.4$, and Probability of failure $=q=0.6$
Now, in 11 steps number of successes $=6$, number of failure $=5$
OR
number of successes $=5$, number of failures $=6$
Required probability $={ }^{11} \mathrm{C}_6 \mathrm{p}^6 \mathrm{q}^5+{ }^{11} \mathrm{C}_5 \mathrm{p}^5 \mathrm{q}^6$.
$\begin{aligned}
& =\frac{11 !}{6 ! 5 !} p^6 q^5+\frac{11 !}{5 ! 6 !} p^5 q^6 \\
& ={ }^{11} C_6 p^5 q^5(p+q) \\
& ={ }^{11} C_6(0-4)^5(0-6)^5(1) \\
& ={ }^{11} C_6(0 \cdot 4)^5(0 \cdot 6)^5 \\
& ={ }^{11} C_6(0 \cdot 24)^5
\end{aligned}$
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