• Think back to replication in an experiment.
• Each replication is what we call a trial.
• We will consider a setting where each trial has two possible outcomes.

Example: suppose you want to know if a coin is fair (both sides equally likely).

• You might flip the coin 100 times (thus running 100 trials).
• Each trial is a flip of the coin with two possible outcomes: heads or tails.

The product of the first \(k\) positive integers \((1, 2, 3, \dots)\) is called k-factorial, denoted \(k!\): \[k! = k \times (k-1) \times\dots\times 3 \times 2 \times 1\] We define \(0!=1\).

Example: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

If \(n\) is a positive integer \((1, 2, 3, \dots)\) and \(x\) is a nonnegative integer \((0, 1, 2, \dots)\) with \(x \le n\), the binomial coefficient is \[\binom{n}{x} = \frac{n!}{x!(n-x)!}\]

Example: \[\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)}\]

What? \[20! = 2,432,902,008,176,640,000\] Your calculator cannot.

Example: \[\binom{20}{17} = \frac{20\times 19\times 18\times 17\times 16\times \dots \times 3\times 2\times 1}{(17\times 16\times \dots \times 3\times 2\times 1)(3\times 2\times 1)}\]

Bernoulli trials are repeated trials of an experiment where:

1. Each trial has two possible outcomes: success and failure.
2. Trials are independent.
3. The probability of success (the success probability) \(p\) remains the same from one trial to the next: \[P(X=\text{success})=p\]

The binomial distribution is the probability distribution for the number of successes in a sequence of Bernoulli trials.

Let \(x\) denote the total number of successes in \(n\) Bernoulli trials with success probability \(p\). The probability distribution of the random variable \(X\) is given by \[P(X=x) = \binom{n}{x}p^x(1-p)^{n-x} \quad\quad x = 0,1,2,\dots,n\] The random variable \(X\) is called a binomial random variable and is said to have the binomial distribution. Because \(n\) and \(p\) fully define this distribution, they are called the distribution’s parameters.

To find a binomial probability formula:

1. Check assumptions.
1. Exactly \(n\) trials to be performed.
2. Two possible outcomes for each trial.
3. Trials are independent (each trial does not impact the result of the next)
4. Success probability \(p\) remains the same from trial to trial.
2. Identify a “success”. Generally, this is whichever of the two possible outcomes we are most interested in.
3. Determine the success probability \(p\).
4. Determine \(n\), the number of trials.
5. Plug \(n\) and \(p\) into the binomial distribution formula.

Consider \(P(X\le x)\).

• We can rewrite this using concepts from the previous chapter \[P(X \le k) = P(X=k \text{ or } X=k-1 \text{ or } \dots \text{ or } X=1 \text{ or } X=0)\]
• Since \(X\) is a discrete random variable, each possible value is disjoint.
• We can use this! \[P(X \le k) = P(X=k) + P(X=k-1) + \dots + P(X=1) + P(X=0)\]

Example Rewrite \(P(X \le 3)\).

We can also extend this concept to work with probabilities like \(P(a < X \le b)\).

Example: \(P(2 < X \le 5)\) What values make up the event of interest?

The shape of a binomial distribution is determined by the success probability:

• If \(p \approx 0.5\), the distribution is approximately symmetric.
• If \(p < 0.5\), the distribution is right-skewed.
• If \(p > 0.5\), the distribution is left-skewed.

• The mean is \(\mu = np\).

• The variance is \(\sigma^2 = np(1-p)\).

Suppose 38.4% of people have dogs. We will take a random sample of 10 people and ask whether they have a dog.

1. Use the binomial distribution to find the probability that exactly 2 of the people in our sample have a dog.

2. Find the probability that between 3 and 5 people (inclusive) have a dog.