CFA Level I Exam · Statistical Distributions for Financial Asset Prices and Returns

Binomial Distribution and Binomial Tree Models for CFA Level I

Q: What is the difference between Bernoulli and binomial distributions?

A Bernoulli distribution describes one trial with two outcomes. A binomial distribution describes the number of successes in n independent Bernoulli trials with the same p. Bernoulli is the special case of binomial with n = 1.

Q: How do I calculate binomial probability on the BA II Plus?

The calculator has no direct binomial function. Use [2nd] [nCr] for the combinations term, then [yx] for the powers, and multiply. For example, 5 [2nd] [nCr] 3 [=] gives 10.

Q: What are the mean and variance of a binomial distribution?

The mean is np and the variance is np(1 − p). The standard deviation is the square root of the variance. These hold when trials are independent and p is constant.

Q: How does a binomial tree model a stock price?

The price moves up by a factor u or down by a factor d each period, with fixed probabilities. In the basic form d = 1/u, so the tree recombines. The number of up moves over n periods follows a binomial distribution.

Updated 6 October 2026 · Fact-checked

A binomial distribution gives the probability of exactly x successes in n independent Bernoulli trials, each with the same success probability p. Use p(x) = n! ÷ [(n − x)! x!] × p^x × (1 − p)^(n − x). A binomial tree applies the same idea to prices, with an up or down move each period.

Understand Binomial Distribution and Binomial Tree Models

A Bernoulli trial is one experiment with two outcomes: success or failure. Success has probability p and failure has probability 1 − p. A Bernoulli random variable takes the value 1 for success and 0 for failure. Its mean is p and its variance is p(1 − p).

A binomial random variable counts the number of successes in n Bernoulli trials. The trials must be independent, and p must be the same in every trial. The count can be any whole number from 0 to n. So the binomial is a discrete distribution. A Bernoulli variable is just a binomial with n = 1.

The probability of exactly x successes has two parts. First, any one sequence with x successes and n − x failures has probability p^x × (1 − p)^(n − x). Second, you count how many such sequences exist. That count is n! ÷ [(n − x)! x!], also written nCx. Multiply the two parts together.

A binomial tree models an asset price as a series of up or down moves. In each period the price is multiplied by an up factor u or a down factor d. With probability p it goes up, and with probability 1 − p it goes down. In the simple version you use, d = 1/u, so an up move followed by a down move returns the price to where it started. This makes the tree recombining. After n periods, the number of up moves follows a binomial distribution, which links the tree to the formula. Binomial trees are also the base for option pricing, which you study under derivatives.

Key formulas to remember

Bernoulli mean and variance: E(X) = p; Var(X) = p(1 − p); X = 1 for success, 0 for failure. One trial only.
Binomial probability: p(x) = n! ÷ [(n − x)! x!] × p^x × (1 − p)^(n − x); Probability of exactly x successes in n independent trials with constant p.
Binomial mean and variance: E(X) = np; Var(X) = np(1 − p); standard deviation = √[np(1 − p)]; Valid under the same conditions: independent trials, constant p.
Number of arrangements: nCx = n! ÷ [(n − x)! x!]; On the BA II Plus: n [2nd] [nCr] x [=].
One-period binomial tree: S(up) = S0 × u; S(down) = S0 × d; d = 1/u; u > 1 and d < 1. Recombining when d = 1/u.
Expected price after one period: E(S1) = p × S0 × u + (1 − p) × S0 × d; A probability-weighted average of the two possible prices. For n periods, use the one-period expected multiplier raised to the power n: E(Sn) = S0 × [p·u + (1 − p)·d]^n.
Cumulative probability: P(X ≤ k) = p(0) + p(1) + ... + p(k); For 'at least' questions, use P(X ≥ k) = 1 − P(X ≤ k − 1).

How to solve Binomial Distribution and Binomial Tree Models questions

Use this method for any binomial or binomial tree question. Most marks are lost by skipping the setup, not by weak arithmetic.

1Check the conditions: two outcomes, fixed number of trials n, independent trials, constant p. If one fails, the binomial does not apply.
2Define what counts as a success and write down n, p and x.
3For a single probability, compute nCx, then p^x, then (1 − p)^(n − x), and multiply the three.
4For 'at most' or 'at least' questions, list the values of x needed. Add their probabilities, or use the complement if that is shorter.
5For mean or variance of a count, use np and np(1 − p). Do not sum probabilities.
6For a tree, label each node with S0 × u^(number of ups) × d^(number of downs). Find each path probability and the number of paths to each node.
7Weight node prices by their probabilities for an expected price.
8Check that your probabilities add to 1, then pick the matching option from the three listed.

Quickest way: Shortcuts for binomial questions under time pressure

When to use it: Use these when the exam gives three numerical options and you have about 90 seconds.

Compute nCx with [2nd] [nCr] on the BA II Plus. Example: 5 [2nd] [nCr] 3 [=] gives 10. On the HP 12C, multiply out the factorial terms by hand.
Use [yx] for powers. Example: 0.6 [yx] 3 [=] gives 0.216.
For mean or variance questions, np and np(1 − p) take seconds. Do them first.
Estimate before computing. If p is small and x is large, the probability should be very small. Cross out any option that is clearly too large.
On a tree, a node reached by one up and one down has two paths, so its probability is 2p(1 − p), not p(1 − p).
Use the one-period expected multiplier raised to the power n: expected price after n periods = S0 × [p·u + (1 − p)·d]^n.

Common mistakes in Binomial Distribution and Binomial Tree Models

Leaving out the nCx term and giving only p^x × (1 − p)^(n − x).
The formula looks like a single sequence, so the number of orderings is forgotten.
Fix: Always write three factors: nCx, p^x and (1 − p)^(n − x). Check that n is on the first factor.
Confusing Bernoulli and binomial distributions.
Both use p and 1 − p.
Fix: Bernoulli is one trial with mean p and variance p(1 − p). Binomial counts successes in n trials with mean np and variance np(1 − p).
Using p for the wrong outcome, for example treating a failure as the success.
The question gives the probability of the opposite event.
Fix: Define success explicitly and set p to match it before using x.
Treating 'at least x' as 'exactly x'.
Rushing past the wording.
Fix: Write P(X ≥ x) = 1 − P(X ≤ x − 1) or add the needed terms.
Forgetting the path count in a multi-period tree, so the middle node probability is wrong.
Students multiply p and 1 − p once and stop.
Fix: Count the paths. Two periods give 1, 2, 1 paths to the top, middle and bottom nodes.
Applying the binomial when trials are not independent or p changes.
The formula is applied automatically.
Fix: Check the conditions first. If p changes between trials or outcomes affect each other, the binomial is not valid.

Worked examples

Example 1

A fund manager's stock picks each beat the benchmark in a year with probability 0.6, independently. Of 5 picks, what is the probability that exactly 3 beat the benchmark? A. 0.2304 B. 0.2592 C. 0.3456

Show the solution

Success is beating the benchmark. n = 5, p = 0.6, x = 3.
nCx = 5! ÷ (2! × 3!) = 10. On the BA II Plus: 5 [2nd] [nCr] 3 [=].
p^x = 0.6³ = 0.216.
(1 − p)^(n − x) = 0.4² = 0.16.
Probability = 10 × 0.216 × 0.16 = 0.3456.
Check the distractors: 0.2304 is the probability of exactly 2, and 0.2592 is the probability of exactly 4.

Answer: C. 0.3456

Example 2

A stock trades at €80. In each period it rises by a factor u = 1.25 with probability 0.55, or falls by a factor d = 0.8 with probability 0.45. What is the expected price after two periods? A. €80.00 B. €87.78 C. €91.20

Show the solution

Two-period prices: Su² = 80 × 1.25 × 1.25 = €125; Sud = 80 × 1.25 × 0.8 = €80; Sd² = 80 × 0.8 × 0.8 = €51.20.
Probabilities: uu = 0.55² = 0.3025; ud or du = 2 × 0.55 × 0.45 = 0.495; dd = 0.45² = 0.2025. These sum to 1.
Expected price = 125 × 0.3025 + 80 × 0.495 + 51.20 × 0.2025.
= 37.8125 + 39.60 + 10.368 = 87.7805.
Check: one-period expected multiplier = 0.55 × 1.25 + 0.45 × 0.8 = 1.0475. Squared is 1.09725625. Multiplied by 80 gives 87.7805.
Option A ignores the drift. It would be the answer only if the expected multiplier were 1.

Answer: B. €87.78

Exam tips

Questions on this topic are often short and numerical. Set up n, p and x in one line before you touch the calculator.
Read 'exactly', 'at most' and 'at least' carefully. They change the answer completely.
On tree questions, count paths to each node. The middle node of a two-period tree has twice the path count of the outer nodes.
Use np and np(1 − p) for mean and variance of a count. Do not compute the full distribution.
Numerical options are listed from smallest to largest. Use that order and a quick estimate to discard two options fast.

Practice questions from Statistical Distributions for Financial Asset Prices and Returns

Binomial Distribution and Binomial Tree Models in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Binomial Distribution and Binomial Tree Models: frequently asked questions

What is the difference between Bernoulli and binomial distributions?

A Bernoulli distribution describes one trial with two outcomes. A binomial distribution describes the number of successes in n independent Bernoulli trials with the same p. Bernoulli is the special case of binomial with n = 1.

How do I calculate binomial probability on the BA II Plus?

The calculator has no direct binomial function. Use [2nd] [nCr] for the combinations term, then [yx] for the powers, and multiply. For example, 5 [2nd] [nCr] 3 [=] gives 10.

What are the mean and variance of a binomial distribution?

The mean is np and the variance is np(1 − p). The standard deviation is the square root of the variance. These hold when trials are independent and p is constant.

How does a binomial tree model a stock price?

The price moves up by a factor u or down by a factor d each period, with fixed probabilities. In the basic form d = 1/u, so the tree recombines. The number of up moves over n periods follows a binomial distribution.