The binomial and geometric distributions – The geometric distribution

Consider repeated independent trials where:

• Each trial has two outcomes (success or failure).
• The probability of success is constant \(p\).
• Trials continue until the first success occurs.

If \(X\) is the number of trials needed to obtain the first success:

\(X \sim \text{Geo}(p)\)

The probability that the first success occurs on the \(r^{th}\) trial is:

\[ P(X=r) = (1-p)^{r-1}p \]

• \(p\) = probability of success
• \(1-p\) = probability of failure
• \(r\) = trial number of the first success

If \(X \sim \text{Geo}(p)\):

• Mean: \( \mu = \frac{1}{p} \)
• Variance: \( \sigma^2 = \frac{1-p}{p^2} \)

A machine produces items where the probability of a defective item is \(p=0.2\).

Let \(X\) be the trial number of the first defective item.

\(X \sim \text{Geo}(0.2)\)

Find \(P(X=4)\).

Step 1: Substitute into the formula

\[ P(X=4) = (1-0.2)^{3}(0.2) \]

Step 2: Calculate

\[ P(X=4) = (0.8)^3(0.2) \]

The probability that the first success occurs after more than \(r\) trials is:

\[ P(X > r) = (1-p)^r \]

Example:

If \(p=0.3\), find \(P(X>3)\).

\[ P(X>3) = (0.7)^3 \]

• Always define the random variable clearly.
• Write the distribution: \(X \sim \text{Geo}(p)\).
• Remember that the geometric distribution counts trials until the first success.
• The mean \(1/p\) represents the expected number of trials.
• Be careful: \(X\) starts at 1, not 0.