Topic 2.10 Notes – The Binomial Distribution

What a Binomial Random Variable Is

A binomial random variable is a discrete random variable that counts successes in a fixed number of trials. If $X \sim \text{Bin}(n,p)$, then:

• $n$ = number of trials
• $p$ = probability of success on each trial
• $X$ can be $0,1,2,\dots,n$

“Success” just means the outcome you’re counting. It could be a made shot, a defective product, or a missed question.

BINS conditions

For a variable to be binomial, all four must hold:

• Binary: each trial is labeled success or failure
• Independent: one trial doesn’t change another
• Number: the number of trials is fixed ahead of time
• Same probability: $p$ stays constant

A couple of easy traps:

• A trial can have several actual outcomes and still be binomial if you group them into success/failure.
  • Example: “uses a credit card” vs “does not use a credit card”
• “Number of successes in the next 12 trials” can be binomial.
• “How many trials until the first success” is not binomial because the number of trials is not fixed.
• Without replacement is not truly independent, but it’s close enough if the sample is no more than 10% of the population.

When you justify binomial on a test, say it in context. Example:

• Let $X$ be the number of defective bulbs in the next 20 bulbs checked.
• Each bulb is defective or not defective.
• The checks are independent.
• There are 20 bulbs, fixed in advance.
• The probability a bulb is defective is $0.08$ each time.

So $X \sim \text{Bin}(20,0.08)$.

Finding Binomial Probabilities

The exact probability of getting exactly $x$ successes is

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Here’s what the pieces mean:

• $\binom{n}{x}$ counts how many ways the $x$ successes can be arranged
• $p^x$ accounts for the successes
• $(1-p)^{n-x}$ accounts for the failures

For example, if $X \sim \text{Bin}(10,0.3)$, this probability model is centered around 3 successes, and the graph helps you see which counts are most likely.

If $X \sim \text{Bin}(10,0.3)$, then

$$P(X=4)=\binom{10}{4}(0.3)^4(0.7)^6$$

Calculator shortcuts matter a lot here:

• $\text{binompdf}(n,p,x)$ gives $P(X=x)$
• $\text{binomcdf}(n,p,x)$ gives $P(X\le x)$

Translating words into probability notation

• exactly $k$ $\rightarrow P(X=k)$
• at most $k$ $\rightarrow P(X\le k)$
• fewer than $k$ $\rightarrow P(X\le k-1)$
• at least $k$ $\rightarrow P(X\ge k)$
• more than $k$ $\rightarrow P(X\ge k+1)$
• between $a$ and $b$, inclusive $\rightarrow P(a\le X\le b)$

Upper-tail probabilities are usually fastest with complements:

$$P(X\ge k)=1-P(X\le k-1)$$

Be careful. $1-P(X\le4)$ equals $P(X\ge5)$, not $P(X\ge4)$.

Mean and Standard Deviation of a Binomial Distribution

For $X \sim \text{Bin}(n,p)$,

$$\mu_X=np$$

$$\sigma_X^2=np(1-p)$$

$$\sigma_X=\sqrt{np(1-p)}$$

If $X \sim \text{Bin}(20,0.08)$, then

• mean $=20(0.08)=1.6$
• standard deviation $=\sqrt{20(0.08)(0.92)}\approx1.21$

Interpret them in context:

• The mean is the long-run average number of defective bulbs in many groups of 20.
• The standard deviation is the typical distance that count is from 1.6 bulbs.

A common mistake is using $np(1-p)$ as the standard deviation. That’s the variance. Also, the mean can be non-integer even though the count itself must be whole numbers.

For fixed $n$, spread is biggest near $p=0.5$ and smaller when $p$ is near 0 or 1.

Estimating Binomial Probabilities with Simulation

Simulation works when you build the same trial structure as the binomial setting.

1. Create a chance model with probability $p$ of success.
2. Run $n$ independent trials.
3. Count successes. That count is one simulated value of $X$.
4. Repeat many times.
5. Use relative frequency to estimate probability.

Here’s what that looks like for $p=0.2$ and $n=5$:

• Let digits 1-2 = success, 3-0 = failure.
• Generate 5 random digits for one repetition.
• If the digits are 1, 7, 2, 9, 4, then $X=2$.
• After many repetitions, estimate $P(X=2)$ by the proportion with exactly 2 successes.

Binomial simulation with random digits

One repetition is the full set of $n$ trials, not one trial. That mistake shows up a lot.