Binomial and Geometric Distributions

Counting successes, and waiting for the first.

12 problems · about 22 minutes · UNC-3.A, UNC-3.B

What this lesson teaches

The student identifies binomial and geometric settings and computes their probabilities and means.

A binomial setting needs binary outcomes, independence, fixed trials and constant p.
The mean is np and the standard deviation is √(np(1 − p)).
A geometric variable counts trials until the first success, with mean 1 ÷ p.

Straightforward practice. Get the method working first.

5 problems

n = 60 and p = 0.5. What is the mean number of successes?
Answer 30
Why 30.
What separates a geometric setting from a binomial one?
Answer The number of trials is not fixed; you wait for the first success.
Why Waiting rather than a fixed count.
n = 100, p = 0.3. Mean number of successes?
Answer 30
Why np.
n = 100, p = 0.5. Standard deviation?
Answer 5
Why √(100 × 0.5 × 0.5).
Geometric with p = 0.25. Expected trials until the first success?
Answer 4
Why 1 ÷ 0.25.

The same ideas with more to keep track of.

3 problems

How many conditions must a binomial setting satisfy?
Answer 4
Why Count them.
4 fair coin tosses. How many patterns have exactly 2 heads?
Answer 6
Why 4C2.
4 fair coin tosses. P(exactly 2 heads), to three decimal places?
Answer 0.375
Why 6 × 0.0625.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Sort each setting by its distribution.
Answer Binomial: Successes in 20 fixed trials, Heads in 10 coin tosses · Geometric: Tosses until the first head, Attempts until the first success
Why A fixed number of trials against waiting for a success.
n = 50, p = 0.2. Mean number of successes?
Answer 10
Why np.
The Mean: n = 40 and p = 0.5. What is the mean number of successes?
Answer 20
Why 20.
The Wait: A geometric setting with p = 0.2. What is the expected number of trials until the first success?
Answer 5
Why 5.