- Distinguish an outcome from the value of a random variable.
- Recognize the assumptions behind a binomial model.
- Use area, rather than point height, for continuous probabilities.
Concepts
A random variable summarizes an outcome
A random variable assigns a number to each outcome. The same experiment can support several variables: number of heads, longest run, or total payoff. A probability mass function assigns weights to discrete values; those weights must be nonnegative and sum to one.
Count successes in a fixed window
A Bernoulli variable is 1 for success and 0 for failure. A binomial count X uses n independent Bernoulli trials with the same success chance p. Then for k=0,…,n. The combination counts placements; the product weights each placement. Changing p across trials or drawing without replacement generally breaks this model.
A density is not a point probability
For a continuous variable, probabilities are areas under its density. A single point has probability zero even when the density there is positive. For a uniform variable on [a,b], the density is 1/(b−a), and an interval inside the support has probability equal to its length divided by b−a. A density may exceed one; its total area must equal one.
Examples
Worked example 1
Two wins in three trials
Three independent trials each succeed with probability 1/2. What is the probability of exactly two successes?
Confirm the model
There is a fixed number of independent trials, and all have the same success probability: X is binomial with n=3 and p=1/2.Choose the successful positions
There are placements: SSF, SFS, and FSS. Each has probability 1/8.Add the weights
The placements are disjoint, so P(X=2)=3/8. A single ordered sequence would give only 1/8.
Worked example 2
Uniform waiting window
A simulated arrival time X is uniform from 2 to 8 minutes. What are P(3≤X≤5) and P(X=4)?
Measure the support
The full interval has length 8−2=6. Its constant density is 1/6 per minute.Measure the event
The interval from 3 to 5 has length 2, so its probability is 2/6=1/3.Separate a point from a neighborhood
P(X=4)=0 because one exact point has zero width. This does not say that times near 4 are impossible.
Explore the idea · unscored
See the distribution, then collect its tail
X counts successes in independent trials with one fixed success chance. Each bar is a binomial probability.
P(X ≥ 6)37.7%
Mean np = 5.00. Standard deviation √[np(1−p)] = 1.58.
Read the probability table
| Successes | Probability | In selected tail |
|---|---|---|
| 0 | 0.0010 | No |
| 1 | 0.0098 | No |
| 2 | 0.0439 | No |
| 3 | 0.1172 | No |
| 4 | 0.2051 | No |
| 5 | 0.2461 | No |
| 6 | 0.2051 | Yes |
| 7 | 0.1172 | Yes |
| 8 | 0.0439 | Yes |
| 9 | 0.0098 | Yes |
| 10 | 0.0010 | Yes |
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Key results
- Identify what is being counted or measured.
- Check distribution assumptions before using its formula.
- A PMF gives point probabilities; a PDF gives density.
- A model must assign total probability one.
Practice problems
Solve related problems or try a timed probability drill.