EN2 - The binomial distribution

EN2 - The binomial distribution

When the same kind of chance experiment is repeated, you can count successes without drawing a huge tree. You will recognise when a binomial model fits, calculate exact probabilities and use a distribution to answer questions about several possible counts. The key is deciding what one trial and one success mean before you reach for the formula.

Recognising binomial trials

If you guess independently on four questions, each with four equally likely answer choices, is getting none correct more likely than getting exactly two correct? There are many ways to be wrong, but also several positions in which the two correct answers could appear. By the end, you will be able to compare those effects exactly.

You already know

Multiply probabilities along one path of a tree, then add probabilities of different paths. A combination counts the ways to choose the successful positions when their order does not matter.

Consider four rolls of an ordinary fair die. Call a six a success, and any other score a failure. The die still has six possible faces, but this question sorts them into just two categories. A success need not be something desirable: a faulty part can be a success for a model that counts faults.

Binomial distribution

A count XX has a binomial distribution, written X∼B(n,p)X\sim\mathrm{B}(n, p), when it counts successes in a fixed number nn of independent trials, each with two categories, success and failure, and the same success probability pp. The possible values are 0,1,…,n0, 1, \ldots, n. Here nn is a positive integer and 0≤p≤10\le p\le1.

For four fair die rolls counting sixes, X∼B(4,16)X\sim\mathrm{B}(4, \frac16). The number of rolls is fixed; one roll does not change another; every roll has success probability 16\frac16. Independence is an assumption about how the trials behave, not a consequence of having two categories.

Contrast drawing four counters from a bag without replacement. Once a green counter is drawn, fewer green counters remain. The next success probability changes, and the trials are dependent, so that count is not binomial. Replacing each counter and mixing before the next draw restores the same chance on each independent draw. Stopping when the first green appears is different again: the number of trials is not fixed in advance.

At the boundary p=0p=0, the count is certainly 00; at p=1p=1, it is certainly nn. Interpret those cases directly. Most calculations use 0<p<10<p<1.

Key idea

Before writing X∼B(n,p)X\sim\mathrm{B}(n, p), identify the fixed trials, the two categories, independence and the unchanged success probability.

The next student's model has the right starting ingredients but misses an important change after each draw.

Calculating exact-count probabilities

The probability of one particular success-and-failure pattern is only one piece of the answer; the combination counts how many patterns give the same total.

For four independent trials with success chance 14\frac14, the pattern success, failure, success, failure has probability (14)2(34)2(\frac14)^2(\frac34)^2. Exactly two successes could occupy any two of the four positions, and (42)=6\binom42=6. All six patterns have the same probability, so multiply by six.

Binomial probability

P(X=x)=(nx)px(1−p)n−x\mathrm{P}(X=x)=\binom nx p^x(1-p)^{n-x}

Use this when X∼B(n,p)X\sim\mathrm{B}(n, p) and xx is one of 0,1,…,n0, 1, \ldots, n. The factor (nx)\binom nx counts the successful positions; pxp^x gives the successes and (1−p)n−x(1-p)^{n-x} gives the failures.
On the formulae sheet.

The formula is the same counting-and-powers structure as a binomial expansion. Indeed, the probabilities of all possible counts are the terms in (p+(1−p))n=1(p+(1-p))^n=1, which explains why they add to one. This probability-sum result follows from the expansion; it is not separately printed on the formulae sheet.

Worked example 1: exactly two successes

Question. A spinner has four equal sectors, one green and three white. It is spun independently four times. Find the probability of exactly two green results.

Plan. Let XX count green results, use n=4n=4, p=14p=\frac14, and multiply one pattern's probability by its number of positions.

X∼B(4,14)fixed independent spinsP(X=2)=(42)(14)2(34)2two successes, two failures=6×116×916=27128evaluate=0.2109375decimal equivalent\begin{aligned} X&\sim\mathrm{B}\left(4, \frac14\right) && \text{fixed independent spins}\\ \mathrm{P}(X=2)&=\binom42\left(\frac14\right)^2\left(\frac34\right)^2 && \text{two successes, two failures}\\ &=6\times\frac1{16}\times\frac9{16}=\frac{27}{128} && \text{evaluate}\\ &=0.2109375 && \text{decimal equivalent} \end{aligned}

Why include (42)\binom42? The question allows the green results in any two positions; it does not specify just green, green, white, white.

Answer. 27128\frac{27}{128}, or 0.2110.211 to three significant figures.

Check. One specified pattern has probability 9256\frac9{256}. Six different patterns give 54256=27128\frac{54}{256}=\frac{27}{128}, a probability between zero and one.

The next example changes only the requested count from two to three. One power increases as the other decreases, while their total remains four.

Worked example 2: exactly three successes

Question. For the same four independent spins, find the probability of exactly three green results.

Plan. Use the same model with x=3x=3, leaving one failure.

P(X=3)=(43)(14)3(34)three successes, one failure=4×164×34four failure positions=364=0.046875evaluate\begin{aligned} \mathrm{P}(X=3)&=\binom43\left(\frac14\right)^3\left(\frac34\right) && \text{three successes, one failure}\\ &=4\times\frac1{64}\times\frac34 && \text{four failure positions}\\ &=\frac3{64}=0.046875 && \text{evaluate} \end{aligned}

Why does the failure have four possible positions? Specifying the one white result determines the other three positions automatically; (43)=(41)=4\binom43=\binom41=4.

Answer. 364\frac3{64}, or 0.04690.0469 to three significant figures.

Check. Three greens should be less likely than two when green has probability only 14\frac14; 0.046875<0.21093750.046875<0.2109375.

Using the same formula for each possible count gives the entire probability distribution. This is a list of all the values the count can take, with the probability of each.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account