Binomial Distribution: Step-by-Step Explanation

The binomial distribution models the number of successes in $n$ independent trials of a binary experiment (success or failure) where the probability of success in each trial is $p$. Let’s break it down step-by-step:

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Step 1: Understand the Problem Setting

• Trial: Each repetition of the experiment (e.g., flipping a coin, rolling a die).
• Success: The desired outcome of interest (e.g., getting heads, rolling a 6).
• Failure: Any outcome that is not the desired one.
• Independent Trials: The outcome of one trial does not influence the outcome of another.

Step 2: Parameters of the Distribution

• $n$: Total number of trials.
• $p$: Probability of success in each trial.
• $q = 1 - p$: Probability of failure.
• $X$: Random variable representing the number of successes in $n$ trials.

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Step 3: Formula for Binomial Probability

The probability of getting exactly $k$ successes ($X = k$) in $n$ trials is given by:

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

Where:

• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$: Binomial coefficient, which counts the number of ways to arrange $k$ successes in $n$ trials.
• $p^k$: Probability of $k$ successes.
• $q^{n-k}$: Probability of $n-k$ failures.

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Step 4: Steps to Calculate Probability

1. Identify $n$, $p$, and $k$:

   • For example, if you flip a coin 5 times ($n = 5$) and want 3 heads ($k = 3$) with $p = 0.5$, these values will be used.

2. Calculate the Binomial Coefficient:

   • Use $\binom{n}{k}$: $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$

3. Compute $p^k$ and $q^{n-k}$:

   • For $p = 0.5$, $q = 1 - 0.5 = 0.5$: $p^k = (0.5)^3 = 0.125, \quad q^{n-k} = (0.5)^2 = 0.25$

4. Combine Using the Formula:

   • Multiply the components: $P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125$

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Step 5: Key Properties

1. Mean (Expected Value): $\mu = n \cdot p$ Example: For $n = 5$, $p = 0.5$: $\mu = 5 \cdot 0.5 = 2.5$.

2. Variance: $\sigma^2 = n \cdot p \cdot q$ Example: For $n = 5$, $p = 0.5$, $q = 0.5$: $\sigma^2 = 5 \cdot 0.5 \cdot 0.5 = 1.25$.

3. Standard Deviation: $\sigma = \sqrt{n \cdot p \cdot q}$

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Step 6: Cumulative Probability

If you want the probability of $X$ being less than or equal to $k$ ($P(X \leq k)$), sum the probabilities for all $j$ from $0$ to $k$:

$P(X \leq k) = \sum_{j=0}^k P(X = j)$

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Applications

• Coin tosses
• Quality control (e.g., defective items in a batch)
• Election polling (e.g., percentage of voters for a candidate)

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Would you like to see a worked example or solve a problem together?

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Relative Questions:

1. How is the binomial distribution related to the normal distribution?
2. What happens to the shape of the distribution as $n$ increases?
3. How do you calculate cumulative binomial probabilities using a calculator?
4. Can you describe the difference between binomial and geometric distributions?
5. What are the assumptions required for a binomial distribution to be valid?

Tip:

If $n$ is large and $p$ is not too close to 0 or 1, the binomial distribution can be approximated by a normal distribution for easier computation.