A man takes a step forward with probability 0.4 and backwards with probability 0.6. Find the probabi

Q: Wait, why do we have to consider two separate cases for $f - b = 1$ and $f - b = -1$?

Great question! The problem asks for the probability of being 'one step away' from the origin. This means the man could be at position $+1$ OR position $-1$. Since these are mutually exclusive events (he cannot be at both places at the same time), we calculate the probability of each scenario independently and add them together.

Q: How do I know if the number of steps $f$ and $b$ will always be integers?

That is a brilliant observation! Notice that for any total number of steps $n$ and displacement $d$, we have $f+b=n$ and $f-b=d$. Adding these gives $2f = n+d$. For $f$ to be an integer, $n+d$ must be even. This is why $n$ and $d$ must have the same parity. If they didn't, the probability would simply be zero!

Q: Can I use the formula for a random walk directly without binomial expansion?

You absolutely can, but the binomial distribution is the 'engine' under the hood of a random walk. By using $\binom{n}{k} p^k q^{n-k}$, you are explicitly accounting for all the different sequences of steps that lead to the same final position. It is the most robust way to ensure you don't miss any combinations.

Q: Why does the term $(0.4 + 0.6)$ become $1$ at the end?

That is the beauty of the algebra! When you factor out the common terms, you are left with the sum of the probabilities of the two cases. Since $p+q=1$ (the sum of all possible outcomes for a single step), that term naturally simplifies, leaving you with a much cleaner calculation.

Q: Is there a shortcut if the number of steps was much larger, say $100$?

For large $n$, the binomial distribution starts to look like a Normal (Gaussian) distribution. You would use the Central Limit Theorem to approximate the probability. But for $n=11$, sticking to the exact binomial method is the safest and most accurate path to the correct answer.

Question

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

Sigma Percentile · Accepted Answer

### The Dance of the Random Walker: A Journey Through Probability

Imagine standing on a long, straight path. You are about to embark on a journey of exactly $11$ steps.

With every move, you flip a mental coin—but it is a biased coin. You are more likely to step backward ($60\%$ chance) than forward ($40\%$ chance).

This is the classic 'Random Walk' problem, a fundamental pillar of statistical physics and probability theory. Today, we are going to break down this journey not just as a math problem, but as a story of paths taken and paths left behind.

### Defining the Terrain

We start with $n = 11$ total steps. Our goal is simple: at the end of these $11$ steps, we want to be exactly $1$ unit away from where we started.

Let $f$ be the number of forward steps and $b$ be the number of backward steps. We know two things immediately: first, the total number of steps is $f + b = 11$.

Second, our net displacement is $f - b$. To be one step away, this displacement must satisfy the condition $|f - b| = 1$. This gives us two distinct realities: either we end at $+1$ or we end at $-1$.

### Solving the Two Realities

Let us look at the first reality: $f - b = 1$. By solving the system of equations $f + b = 11$ and $f - b = 1$, we add them to get $2f = 12$.

This tells us we must take exactly $f = 6$ forward steps and $b = 5$ backward steps.

Now, consider the second reality: $f - b = -1$. Solving $f + b = 11$ and $f - b = -1$ gives us $2f = 10$, meaning we need $f = 5$ forward steps and $b = 6$ backward steps.

These are the only two ways to satisfy the condition. Any other combination of steps would land us at a displacement of $3, 5, 7, 9,$ or $11$, but never $1$.

### The Power of the Binomial Distribution

Now, we bring in the heavy artillery: the Binomial Distribution. The probability of taking $k$ forward steps in $n$ trials is given by:

$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$

Here, $p = 0.4$ and $q = 0.6$.

For our first case ($f=6$), the probability is:

$$P(X=6) = \binom{11}{6} (0.4)^6 (0.6)^5$$

For our second case ($f=5$), the probability is:

$$P(X=5) = \binom{11}{5} (0.4)^5 (0.6)^6$$

We are summing these two probabilities:

$$P_{total} = \binom{11}{6} (0.4)^6 (0.6)^5 + \binom{11}{5} (0.4)^5 (0.6)^6$$

### The Elegance of Symmetry

Here is where the magic happens. Recall that $\binom{11}{6}$ is identical to $\binom{11}{5}$, which is $462$.

We can factor out the common terms:

$$\binom{11}{5} (0.4)^5 (0.6)^5 [0.4 + 0.6]$$

Since $0.4 + 0.6 = 1$, the entire expression collapses into a much simpler form:

$$462 	imes (0.24)^5$$

Calculating this, we arrive at approximately **$0.37$**.

### The Takeaway

What have we learned? We learned that in a random walk, the parity of the steps matters.

If you take an odd number of steps, you can never return to the origin (displacement $0$), because $n$ and $d$ must have the same parity.

This problem is a beautiful reminder that even in the chaos of random movement, there is a rigid, underlying structure governed by the laws of combinatorics. Keep practicing, keep visualizing, and remember: every complex problem is just a collection of simple, logical steps waiting to be connected.

Animated Solution for Mathematics - Probability: A man takes a step forward with probability 0.4 and backwards with probability 0.6. Find the probability that at the end of eleven steps he is one step away from the starting point.

Enter Numerical Value:

Visualized Solution

Visualizing the Random Walk

Step Probabilities

Defining the Variables

The Displacement Constraint

Splitting into Cases

Solving for Case 1: $f - b = 1$

Solving for Case 2: $f - b = - 1$

The Binomial Probability Formula

Setting up Case 1 Probability

Setting up Case 2 Probability

Summing the Probabilities

Factoring the Expression

The Final Calculation

Summary and Way Forward

The Sigma Insight: Binomial Distribution

The Dance of the Random Walker

Defining the Terrain

Solving the Two Realities

The Power of the Binomial Distribution

The Elegance of Symmetry

The Takeaway

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