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Counts built from repeated identical independent trials, why the binomial carries a coefficient and the geometric does not, and which count a story wants.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will check whether a story really supplies independent identical two-outcome trials, decide whether it fixes the number of trials in advance, and use the binomial or the geometric mass function accordingly. You will also say why the binomial carries a coefficient counting arrangements and the geometric does not.
The binomial coefficient came from the counting lesson and the product of independent probabilities from the independence lesson. This lesson is those two facts multiplied together, once, and then given a name — which is why the formula is worth deriving rather than memorising.
A Bernoulli trial is one attempt with two outcomes, success with probability $p$ and failure with probability $1 - p$. A binomial variable counts the successes in $n$ such trials, fixed in advance, independent, all with the same $p$; written $X \sim \mathrm{Bin}(n, p)$. A geometric variable counts the trials up to and including the first success. The negative binomial waits for the $r$-th success, and the geometric is its $r = 1$ case.
Start with a single trial: success with probability $p$, failure with probability $1 - p$. Repeat it independently, always with the same $p$, and two different questions produce the two families of this lesson.
How many succeeded, in $n$ trials? That is binomial:
$$P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k}.$$
The formula is two things multiplied. $p^{k}(1-p)^{n-k}$ is what one particular sequence with $k$ successes is worth; $\binom{n}{k}$ is how many such sequences there are. Every binomial calculation is that pair.
How long until the first success? That is geometric:
$$P(Y = k) = (1-p)^{k-1} p.$$
Here there is no coefficient at all, because first success fixes the order completely: everything before trial $k$ failed. That absence is the clearest sign of which family you are in.
Both rest on the same three conditions, and all three are worth checking out loud: the trials are independent, they have the same $p$, and each is a plain success or failure. Sampling without replacement breaks the first two, which is why drawing cards is neither.
Another way: table
Four questions about the same coin, tossed with $p = 0.5$.
| Question | Family | Answer |
|---|---|---|
| Is this toss heads? | Bernoulli | $0.5$ |
| How many heads in $4$ tosses? | Binomial | $\binom{4}{k}/16$ |
| Which toss is the first head? | Geometric | $0.5^{k}$ |
| Which toss is the third head? | Negative binomial | needs a coefficient again |
Another way: steps
Three trials with $p = 0.2$, written out in full.
| Event | Sequences | Each worth | Probability |
|---|---|---|---|
| Exactly one success | SFF, FSF, FFS | $0.2 \times 0.8^{2} = 0.128$ | $3 \times 0.128 = 0.384$ |
| First success on trial $3$ | FFS only | $0.8^{2} \times 0.2 = 0.128$ | $0.128$ |
The same $0.128$ appears in both rows. The difference is entirely in how many sequences the description covers: three for the binomial question, one for the geometric. If you can say how many sequences your event covers, you have the formula whether or not you remember it.
Using the binomial for sampling without replacement. The trials are then neither independent nor identical. The right model is the hypergeometric; the binomial is a decent approximation only when the sample is a small part of the population.
Putting a binomial coefficient in a geometric probability. First success on trial $k$ has exactly one arrangement.
Forgetting that $p$ must be the same every trial. A machine that wears out as it runs does not produce binomial counts.
*Reading at least one as a single term.* $P(X \ge 1) = 1 - (1-p)^{n}$, which is one line; adding the other terms is $n$ lines and the same answer.
Five independent attempts, each succeeding with probability $0.4$. What is $P(X = 2)$?
Check the three conditions first.
One sequence with two successes is worth $0.4^{2} \times 0.6^{3} = 0.03456$, and there are $\binom{5}{2} = 10$ of them.
The two factors.
So $P(X = 2) = 0.3456$.
Multiply.
The same attempts. What is the probability the first success is on attempt $4$?
The run is not fixed in advance.
Three failures then a success: $0.6^{3} \times 0.4 = 0.0864$.
One arrangement only.
And $P(Y > 4) = 0.6^{4} = 0.1296$: no success in the first four.
The tail is a single product.
The trials are independent with the same $p = 0.5$ and the run is fixed at four, so this is binomial.
Check before computing.
One sequence with three heads is worth $0.5^{4} = 1/16$, and there are $\binom{4}{3} = 4$ of them.
So the probability is $4/16 = 0.25$.
A machine produces a good part with probability $1/5$, independently each time. Three parts are made and $X$ counts the good ones. Give the mass function of $X$.
| Probability | |
|---|---|
| $P(X = 0)$ | |
| $P(X = 1)$ | |
| $P(X = 2)$ | |
| $P(X = 3)$ |
Which model fits the number of correct answers when twelve questions are all guessed?
Four independent attempts each succeed with probability $3/10$. Give the probability of exactly two successes, and the probability of four.
Exactly two: p. All four: q.
Each attempt succeeds with probability $3/5$, independently. Put these four events in order, least likely first.
Number the steps in order (write the number in the box):
Each attempt succeeds with probability $3/5$, independently. What is the probability that the first success is on attempt $2$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each family to the quantity it counts, for trials that each succeed with probability $p$.
| The outcome of one trial, as $1$ or $0$ | The number of successes in $17$ trials fixed in advance | The number of trials up to and including the first success | The number of trials up to and including the $r$-th success | |
|---|---|---|---|---|
| Bernoulli | ||||
| Binomial | ||||
| Geometric | ||||
| Negative binomial |
You can check the three conditions for repeated trials, choose between the binomial and the geometric, and compute either mass function. Say in your own words why sampling without replacement fits neither of them.
9. Your turn: four independent tosses of a fair coin; find the probability of exactly three heads, step 3