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B. Probability Distributions · Discrete Probability Distributions

Binomial Experiments and the Binomial Probability Formula

B1.4About 75 minutes
Learning goals
  • I can identify the four conditions of a binomial experiment and decide whether a situation is binomial.
  • I can derive the binomial probability formula from a tree diagram.
  • I can use P(X = k) = C(n, k)pᵏqⁿ⁻ᵏ to find binomial probabilities.
  • I can connect binomial probabilities to Pascal's triangle and the binomial expansion of (q + p)ⁿ.

Words to know

Bernoulli trial
A single trial with exactly two outcomes, called success (probability p) and failure (probability q = 1 − p).
binomial experiment
A fixed number n of independent Bernoulli trials, each with the same probability of success p.
binomial random variable
X = the number of successes in a binomial experiment. We write X ~ B(n, p).
independent trials
Trials where the result of one trial does not change the probabilities for the others (for example, sampling with replacement).
C(n, k)
The number of ways to choose k items from n, n!/(k!(n − k)!). Also written ₙCₖ or (n k) stacked in brackets.

Repeat the same chance, many times

A basketball player makes 70% of her free throws. She takes 5 shots. What is the probability she makes exactly 4?

A student guesses on all 10 questions of a multiple-choice quiz with 4 choices each. What is the probability of exactly 3 right?

A factory makes light bulbs, and 0.5% are defective. In a pack of 4, what is the probability that none is defective?

These problems have the same structure. Something with two outcomes is repeated a fixed number of times, with the same chance each time, and we count the successes. This is a binomial experiment.

The four conditions

A random variable X is binomial when all four are true:

  1. Fixed number of trials, n.
  2. Two outcomes per trial: success or failure. ("Success" just means the outcome we are counting; it can be something bad, like a defective bulb.)
  3. Independent trials: one trial does not affect another.
  4. Constant probability of success p on every trial. Then q = 1 − p is the probability of failure.

X counts the number of successes, so X can be 0, 1, 2, ..., n. We write X ~ B(n, p).

A single trial with two outcomes is called a Bernoulli trial, after the Swiss mathematician Jacob Bernoulli.

Worked example

Example 1: Binomial or not?

Decide whether X is binomial. If it is, give n and p. (a) X = number of 6s in 10 rolls of a die. (b) X = number of hearts when 5 cards are dealt from one deck. (c) X = number of rolls until the first 6. (d) X = number of correct guesses on 20 true-or-false questions.

  1. (a) Fixed n = 10, success = rolling a 6, independent rolls, p = 1/6 each time. Binomial, B(10, 1/6).
  2. (b) Dealing without replacement: after a heart is dealt, fewer hearts remain, so p changes and the trials are dependent. Not binomial (it is hypergeometric, lesson 8).
  3. (c) The number of trials is not fixed. Not binomial.
  4. (d) n = 20, success = correct guess, p = 1/2, independent. Binomial, B(20, 0.5).

(a) and (d) are binomial; (b) and (c) are not.

Building the formula with a tree

Think of 3 trials. Each trial has a success S (probability p) or a failure F (probability q). Because the trials are independent, we multiply along each path of the tree.

pSqFpSqFpSqFpSqFpSqFpSqFpSqFoutcomeprobabilityXSSSp³3SSFp²q2SFSp²q2SFFpq²1FSSp²q2FSFpq²1FFSpq²1FFFq³03 paths have X = 2, each with probability p²q, so P(X = 2) = 3p²q = C(3, 2)p²q
Every path with exactly k successes has probability pᵏqⁿ⁻ᵏ. The only question is how many such paths there are.

For X = 2 in 3 trials, the paths are SSF, SFS and FSS. Each has probability p·p·q = p²q, and there are 3 of them, so P(X = 2) = 3p²q.

In general, a path with exactly k successes in n trials has probability pᵏqⁿ⁻ᵏ. The number of such paths is the number of ways to choose which k of the n trials are successes: C(n, k). So

P(X = k) = C(n, k) pᵏ qⁿ⁻ᵏ, k = 0, 1, ..., n

where C(n, k) = n!/(k!(n − k)!), as in Permutations and Combinations.

n = 01n = 111n = 2121n = 31331n = 414641n = 515101051Row 5 gives C(5, k) for k = 0 to 5: the number of ways to place k successes among 5 trials
The coefficients C(n, k) are the entries of Pascal's triangle. For n = 3 they are 1, 3, 3, 1, matching the 1, 3, 3, 1 paths with 0, 1, 2, 3 successes in the tree.

Remember from Grade 11

In Sequences and Series (MCR3U) you expanded (a + b)ⁿ with Pascal's triangle. Expand (q + p)ⁿ and each term C(n, k)pᵏqⁿ⁻ᵏ is one binomial probability. Since q + p = 1, the terms add to 1ⁿ = 1. That is why it is called the binomial distribution.

Worked example

Example 2: Free throws

A player makes 70% of her free throws. Assuming independent shots, find the probability that she makes exactly 4 of her next 5.

  1. Success = make, n = 5, p = 0.7, q = 0.3, k = 4.
  2. P(X = 4) = C(5, 4)(0.7)⁴(0.3)¹.
  3. C(5, 4) = 5 and 0.7⁴ = 0.2401.
  4. P(X = 4) = 5 × 0.2401 × 0.3 = 0.36015.

P(X = 4) ≈ 0.360, about a 36% chance.

Worked example

Example 3: Guessing on a quiz

A quiz has 10 multiple-choice questions, each with 4 choices. A student guesses every answer. Find the probability of (a) exactly 3 correct, (b) no correct answers, (c) at least 1 correct.

  1. n = 10, p = 1/4 = 0.25, q = 0.75.
  2. (a) P(X = 3) = C(10, 3)(0.25)³(0.75)⁷ = 120 × 0.015625 × 0.13348... ≈ 0.2503.
  3. (b) P(X = 0) = (0.75)¹⁰ ≈ 0.0563.
  4. (c) "At least 1" is the complement of "none": P(X ≥ 1) = 1 − P(X = 0) ≈ 1 − 0.0563 = 0.9437.

(a) ≈ 0.250; (b) ≈ 0.056; (c) ≈ 0.944

"At least one" means use the complement

P(X ≥ 1) = 1 − P(X = 0) = 1 − qⁿ. This is much faster than adding P(1) + P(2) + ... + P(n).

Worked example

Example 4: At least one 6

You roll a die 4 times. What is the probability of at least one 6? Is it more or less than 1/2?

  1. X ~ B(4, 1/6). P(X = 0) = (5/6)⁴ = 625/1296 ≈ 0.4823.
  2. P(X ≥ 1) = 1 − 0.4823 = 0.5177.

About 0.518, a little more than 1/2.

Just for fun

In statistics, a factory that finds a defective light bulb may record it as a success. It is the only job where finding a broken bulb counts as a win.

Hands-on activity

Coins, dice and the formula

You need: 5 coins (or one coin tossed 5 times), one die, pencil, paper and a calculator.

  1. Toss 5 coins and count heads. Repeat 32 times, tallying the number of heads (0 to 5).
  2. Use the formula with n = 5, p = 0.5 to find P(X = k) for k = 0 to 5. (You should get 1/32, 5/32, 10/32, 10/32, 5/32, 1/32: row 5 of Pascal's triangle over 2⁵.) Multiply by 32 to get expected counts: 1, 5, 10, 10, 5, 1.
  3. Compare your counts with the expected counts.
  4. Now roll the die 5 times, count the 6s (n = 5, p = 1/6), and repeat 30 times. Before you start, use the formula to predict how often you will get no 6s at all.

What to look for: for the coins, results such as 0, 6, 9, 11, 5, 1 are typical. For the die, P(no 6s) = (5/6)⁵ ≈ 0.402, so about 12 of your 30 sets should have no 6s.

Your turn

1

X ~ B(6, 0.5). Find P(X = 3).

2

X ~ B(8, 0.2). Find P(X = 0) and P(X ≥ 1).

3

A die is rolled 6 times. Find the probability of exactly two 6s.

4

Is this binomial? "X = number of red cards in 4 draws from a deck, replacing and shuffling each time." If yes, give n and p.

5

Is this binomial? "X = number of students absent from a class of 30 on a day when a flu is going around."

6

In 12 independent trials with p = 0.1, find P(X = 2).

7

Write out all the terms of (q + p)⁴ and say which one equals P(X = 2) for X ~ B(4, p).

More practice: warm-up

1

How many outcomes does each trial of a binomial experiment have?

2

If p = 0.35, what is q?

3

Evaluate C(5, 2).

4

X ~ B(3, 0.5). Find P(X = 3).

5

In X ~ B(n, p), what does n stand for?

More practice: on your own

1

X ~ B(4, 0.6). Find P(X = 2).

2

About 5% of eggs in a large shipment are cracked. In a carton of 20, find the probability that exactly 1 is cracked.

3

A spinner lands on blue with probability 0.4. In 7 spins, find P(at least one blue).

4

Guessing on 10 four-choice questions, find P(at least 5 correct). Use P(5) ≈ 0.0584, P(6) ≈ 0.0162, P(7) ≈ 0.0031, P(8) ≈ 0.0004 and P(9) + P(10) ≈ 0.0000.

5

Explain why drawing 3 cards without replacement and counting aces is not binomial, but becomes close to binomial if the cards are drawn from 20 shuffled decks.

Challenge

1

Show that P(X = k) for X ~ B(n, p) add to 1, using the binomial theorem.

2

A test has 15 multiple-choice questions with 5 choices each. What is the probability of getting exactly 3 correct by guessing? How does this compare with the most likely score?

3

How many times must you roll a die so that the probability of at least one 6 is more than 0.9?

Check your understanding

Quiz

1.

Which is not a condition of a binomial experiment?

2.

X ~ B(5, 0.7). What is P(X = 4)?

3.

Which situation is binomial?

4.

X ~ B(10, 0.25). What is P(X ≥ 1)?

5.

For X ~ B(4, p), the number of paths in a tree diagram with exactly 2 successes is...

6.

In a binomial experiment, a "success" is...

For parents and teachers

Big idea (B1.4): a binomial random variable counts successes in n independent trials with the same probability p. Its probabilities are P(X = k) = C(n, k)pᵏ(1 − p)ⁿ⁻ᵏ. The formula comes from a tree diagram: each path with k successes has probability pᵏqⁿ⁻ᵏ, and there are C(n, k) such paths.

How to help: give your teen everyday situations (free throws, guessing on a quiz, drawing cards with or without putting them back) and ask whether each one is binomial and why. Checking the four conditions is the most important skill here.

Common difficulties:

  • Forgetting the C(n, k) factor (using only pᵏqⁿ⁻ᵏ, which is the probability of one particular path).
  • Using the formula when the trials are dependent (drawing without replacement).
  • Computing "at least one" by adding many terms instead of using 1 − qⁿ.
  • Calculator slips with powers: enter (0.75)^7, not 0.75 × 7.