Skill: Model repeated trials with binomial distributions

A binomial distribution is useful when:

• There is a fixed number of trials, $n$.
• Each trial has only two outcomes: success or failure.
• The trials are independent.
• The probability of success, $p$, stays the same each time.

We write this as

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

where $X$ is the number of successes.

Example

A basketball player makes a free throw with probability $0.7$. Suppose the player takes $10$ free throws. What is the probability that the player makes exactly $7$ shots?

Step 1: Check the binomial conditions

Each free throw is one trial.

• There are a fixed number of trials: $n=10$.
• A success means making the shot; a failure means missing it.
• Assume the shots are independent.
• The probability of success is constant: $p=0.7$.

Therefore,

$$X\sim\operatorname{Bin}(10,0.7),$$

where $X$ represents the number of shots made.

Step 2: Use the binomial probability formula

The probability of exactly $k$ successes is

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

For exactly $7$ successes, substitute $n=10$, $k=7$, and $p=0.7$:

$$P(X=7)=\binom{10}{7}(0.7)^7(0.3)^3.$$

The combination $\binom{10}{7}$ counts the different ways that $7$ of the $10$ shots could be made.

$$P(X=7)=120(0.7)^7(0.3)^3\approx 0.2668.$$

So, the probability that the player makes exactly $7$ shots is approximately

$$\boxed{0.267}$$

or about $\boxed{26.7\%}$.

A probability table or distribution graph for $X\sim\operatorname{Bin}(10,0.7)$ would list or display probabilities for $X=0,1,2,\ldots,10$. The value for $X=7$ would be about $0.267$.

Step 3: Find the expected value and variability

The expected number of successes is

$$E(X)=np=10(0.7)=7.$$

So, over many groups of $10$ shots, the player would make about $7$ shots on average.

The variance, a measure of variability, is

$$\operatorname{Var}(X)=np(1-p)=10(0.7)(0.3)=2.1.$$

Thus, this binomial model describes both the likely number of makes and how much that number tends to vary from one group of $10$ shots to another.