Binomial Situations

Counting successes in a fixed number of tries.

12 problems · about 22 minutes · TEKS P.S.2.G, S-MD.A.3

What this lesson teaches

The student recognises binomial settings and computes binomial probabilities and means.

A binomial setting needs fixed trials, two outcomes, constant p, and independence.
P(exactly k) = nCk × pᵏ × (1 − p)^(n − k).
The mean of a binomial variable is np.

Straightforward practice. Get the method working first.

5 problems

n = 60, p = 0.5. What is the mean number of successes?
Answer 30
Why 30.
Why is drawing cards without replacement not binomial?
Answer The trials are not independent, since the deck changes.
Why Independence fails.
n = 20, p = 0.5. Mean number of successes?
Answer 10
Why np.
n = 50, p = 0.2. Mean?
Answer 10
Why 50 × 0.2.
5 tosses. How many patterns have exactly 2 heads?
Answer 10
Why 5C2.

The same ideas with more to keep track of.

3 problems

4 tosses of a fair coin. P(exactly 2 heads), as a decimal to three places?
Answer 0.375
Why 6 × 0.0625.
How many conditions must a binomial setting satisfy?
Answer 4
Why Count them.
n = 100, p = 0.3. Mean?
Answer 30
Why 100 × 0.3.

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

4 problems

Which settings are binomial?
Answer Tossing a coin 10 times and counting heads; Asking 50 people a yes-or-no question
Why One breaks the fixed count and one breaks independence.
n = 12, p = 0.25. Mean?
Answer 3
Why 12 × 0.25.
The Mean: 40 tosses of a fair coin. What is the mean number of heads?
Answer 20
Why 20.
The Patterns: 4 coin tosses. How many patterns have exactly 2 heads?
Answer 6
Why 6.