In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 1/3

Q: Wait, why do we reject values outside the range $4 \le x \le 6$?

The inequality $|x - y| \le 2$ acts as a filter. If $x$ were $3$, then $y$ would be $7$, and $|3 - 7| = 4$, which violates our condition. The math forces us to stay within the 'balanced' zone. Always trust the inequality to define your boundaries!

Q: Why can't we just use simple probability?

Simple probability works for single events, but here we have a sequence of 10 independent trials. The Binomial Distribution is the only tool that accounts for the specific number of ways to arrange these wins and losses. It is the bridge between individual match outcomes and the tournament's aggregate result.

Q: Is there a shortcut for the binomial coefficients?

Yes! Notice the symmetry in Pascal's Triangle. $\binom{10}{4}$ is the same as $\binom{10}{6}$. Recognizing these patterns is what separates a good student from a great one. It saves time and reduces the chance of calculation errors.

Question

In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $P(|x - y| \le 2)$ is $p$, then $3^9 p$ equals ______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We define $x$ as the number of wins and $y$ as the number of losses in a series of $10$ matches. Given the total number of matches, we establish the fundamental equation:

$$x + y = 10$$

The problem imposes a constraint on the "balance" of the team, defined by the absolute difference between wins and losses:

$$|x - y| \le 2$$

### The Geometry of Constraints

To solve this, we substitute $y = 10 - x$ into the constraint to reduce the system to a single variable:

$$|x - (10 - x)| \le 2 \implies |2x - 10| \le 2$$

Expanding the modulus, we solve the inequality:

$$-2 \le 2x - 10 \le 2$$

Adding $10$ to all sides yields $8 \le 2x \le 12$, which simplifies to:

$$4 \le x \le 6$$

This confirms that the team must win exactly $4$, $5$, or $6$ matches to satisfy the condition.

### The Binomial Engine

Since each match is an independent trial, we utilize the Binomial Distribution formula:

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

Here, $n=10$, $p = \frac{1}{3}$, and $q = \frac{2}{3}$. We seek the total probability $P = P(x=4) + P(x=5) + P(x=6)$:

$$P = \binom{10}{4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6 + \binom{10}{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5 + \binom{10}{6}\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4$$

Factoring out the common denominator $3^{10}$, we focus on the numerators:

$$P = \frac{1}{3^{10}} \left[ \binom{10}{4} \cdot 2^6 + \binom{10}{5} \cdot 2^5 + \binom{10}{6} \cdot 2^4 \right]$$

### Final Calculation

We evaluate the binomial coefficients and powers of $2$:
$\binom{10}{4} = 210$, $\binom{10}{5} = 252$, and $\binom{10}{6} = 210$.
$2^6 = 64$, $2^5 = 32$, and $2^4 = 16$.

Substituting these values into the expression:

$$P = \frac{210 \cdot 64 + 252 \cdot 32 + 210 \cdot 16}{3^{10}}$$

$$P = \frac{13440 + 8064 + 3360}{3^{10}} = \frac{24864}{3^{10}}$$

To find the final requested value, we multiply $P$ by $3^9$:

$$3^9 \cdot P = 3^9 \cdot \frac{24864}{3^{10}} = \frac{24864}{3}$$

Performing the division, we arrive at the final result:

**8288**

Enter Numerical Value:

Visualized Solution

Defining the Variables

The Probability Condition

Solving the Inequality

Identifying Possible Wins

Applying Binomial Distribution

Simplifying the Expression

Atomic Calculations

Final Summation

The Final Answer

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Geometry of Constraints

The Binomial Engine

Final Calculation

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