A binomial setting counts successes in a fixed number of independent trials, each with the same probability of success.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The four conditions
A fixed number of trials, two outcomes per trial, the same probability each time, and independent trials.
When it does not apply
Drawing without replacement breaks independence, and waiting for the first success breaks the fixed count.
The formula
P(exactly k successes) = nCk × pᵏ × (1 − p)^(n − k).
What each part does
The combination counts which trials succeeded, and the powers give the probability of one particular such pattern.
The mean
The mean of a binomial variable is np. Forty tosses of a fair coin averages 20 heads.
Four conditions
Fixed number of trials, two outcomes each, constant success probability, and independent trials. All four are needed, and real situations frequently violate one — usually independence.
The formula and where it comes from
The probability of exactly k successes multiplies the probability of one such sequence by the number of sequences, which is a combination. The combination is doing the counting and the powers the probability.
Mean and standard deviation
The mean is np and the standard deviation √(np(1 − p)). The mean is intuitive — 100 flips of a fair coin average 50 heads — and the standard deviation says how far from 50 is unremarkable.
"At least" usually means complement
The probability of at least one success is easier as 1 minus the probability of none, rather than summing every case from one upwards. Recognising that shortcut saves most of the work.
Step 2: Try It Yourself
Tap and try it out.
0 heads
1 head
2 heads
3 heads
2 heads has the most. It has 5 more than 0 heads.
Step 3: Watch an Example
One step at a time.
Watch Sana Compute a Binomial Probability
Sana tosses a fair coin 4 times and wants exactly 2 heads.
- Step 1
All four conditions hold: 4 trials, two outcomes, p = 0.5 throughout, and independent tosses.
Step 4: Your Turn
Practice makes it stick.
The Mean
Problem 1 of 2
40 tosses of a fair coin. What is the mean number of heads?
The Patterns
Problem 2 of 2
4 coin tosses. How many patterns have exactly 2 heads?
Count the Successes
1 of 8
n = 20, p = 0.5. Mean number of successes?
2 of 8
n = 50, p = 0.2. Mean?
3 of 8
5 tosses. How many patterns have exactly 2 heads?
4 of 8
4 tosses of a fair coin. P(exactly 2 heads), as a decimal to three places?
5 of 8
How many conditions must a binomial setting satisfy?
6 of 8
n = 100, p = 0.3. Mean?
7 of 8
Which settings are binomial?
8 of 8
n = 12, p = 0.25. Mean?
Step 5: Quick Check
Show what you know.
Question 1 of 2
n = 60, p = 0.5. What is the mean number of successes?
Question 2 of 2
Why is drawing cards without replacement not binomial?
What You Learned
- 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.