Binomial Distribution

A basketball player takes $5$ free throws. A quiz has $10$ multiple-choice questions you guess on. A factory tests $20$ light bulbs. Each of these repeats the same yes-or-no trial several times and counts the “yes” results. The binomial distribution gives the probability of every possible count, with one formula.

Key ideas

When is it binomial?

A random variable $X$ has a binomial distribution when all four conditions hold:

1. There is a fixed number of trials, $n$.
2. Each trial has two outcomes: success or failure.
3. The probability of success, $p$, is the same on every trial.
4. The trials are independent.

$X$ is the number of successes in the $n$ trials, so $x = 0, 1, 2, \dots, n$. The probability of failure is $1 - p$ (sometimes written $q$).

“Success” just means the outcome you’re counting. It can be something bad, like a defective bulb.

The formula

$$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \qquad k = 0, 1, 2, \dots, n$$

Here’s why each part is there:

• $p^k (1 - p)^{n - k}$ is the probability of one particular order with $k$ successes and $n - k$ failures (multiply, because the trials are independent).
• $\binom{n}{k}$ counts how many orders there are: the number of ways to choose which $k$ of the $n$ trials are the successes. (See combinations; it’s also written $C(n, k)$ or ${}_nC_k$.)

Technology: on a TI-83/84, binompdf(n, p, k) gives $P(X = k)$, and binomcdf(n, p, k) gives $P(X \le k)$. In a spreadsheet, use =BINOM.DIST(k, n, p, FALSE) and =BINOM.DIST(k, n, p, TRUE).

Expected value

For a binomial distribution,

$$E(X) = np$$

This makes sense: if you take $10$ shots with a $30\%$ chance each, you expect about $10 \times 0.3 = 3$ to go in.

The table and histogram

The distribution for $n = 10$ and $p = 0.3$, rounded to $4$ decimal places:

$k$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$P(X = k)$	$0.0282$	$0.1211$	$0.2335$	$0.2668$	$0.2001$	$0.1029$	$0.0368$	$0.0090$	$0.0014$	$0.0001$	$0.0000$

(The rounded values add to $0.9999$; the exact values add to $1$.)

The binomial distribution with $n = 10$ and $p = 0.3$. The peak is at the expected value $np = 3$.

How the shape changes as n increases

When $p = 0.5$, the histogram is symmetric. When $p$ is less than $0.5$, it’s skewed right (a long tail toward the large values); when $p$ is greater than $0.5$, it’s skewed left.

As $n$ increases (with $p$ fixed), the histogram:

• moves to the right, because its centre $np$ grows;
• spreads out over more values;
• becomes more symmetric and bell-shaped, even if $p$ isn’t $0.5$.

That bell shape is the reason a binomial distribution can be approximated by a normal distribution when $n$ is large; see normal approximation.

Binomial distributions with $p = 0.3$ and $n = 5$, $15$, and $40$. As $n$ grows, the shape becomes more symmetric.

“At least” problems

For “at least one”, use the complement: the opposite of “at least one success” is “no successes”.

$$P(X \ge 1) = 1 - P(X = 0)$$

Similarly, $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$. This is much faster than adding up many terms.

Worked examples

Example 1: Is it binomial?

Decide whether $X$ is binomial. If it is, give $n$ and $p$.

• (a) A die is rolled $12$ times, and $X$ is the number of $5$s.
• (b) Five cards are dealt from a deck, and $X$ is the number of hearts.
• (c) A coin is flipped until it lands heads, and $X$ is the number of flips.

Solution.

(a) Yes. There are $12$ independent trials, each a success ($5$) or failure, with $p = \tfrac{1}{6}$ every time. So $n = 12$, $p = \tfrac{1}{6}$.

(b) No. The cards aren’t replaced, so the trials are dependent: the chance of a heart changes after each card. (This is the hypergeometric distribution.)

(c) No. The number of trials isn’t fixed.

Example 2: Free throws

A basketball player makes $70\%$ of her free throws. She takes $5$ shots, and the shots are independent. Find the probability that she makes exactly $4$.

Solution. This is binomial with $n = 5$, $p = 0.7$, and $k = 4$.

$$\begin{aligned} P(X = 4) &= \binom{5}{4} (0.7)^4 (0.3)^1 \\ &= 5 \times 0.2401 \times 0.3 \\ &= 0.36015 \approx 0.3602 \end{aligned}$$

The $\binom{5}{4} = 5$ counts which shot is the miss: the first, second, third, fourth, or fifth.

Example 3: Using the table

For $n = 10$ and $p = 0.3$, use the table above to find $P(X \le 2)$ and $E(X)$.

Solution.

$$P(X \le 2) = P(0) + P(1) + P(2) \approx 0.0282 + 0.1211 + 0.2335 = 0.3828$$ $$E(X) = np = 10(0.3) = 3$$

On the histogram, $3$ is exactly where the tallest bar is.

Example 4: At least one defective

A large shipment of light bulbs has $4\%$ defective. A sample of $20$ bulbs is tested. (The shipment is so large that the trials are close enough to independent.) Find the probability that at least one bulb is defective, and that at least two are.

Solution. This is binomial with $n = 20$ and $p = 0.04$. Use the complement.

$$P(X = 0) = (0.96)^{20} \approx 0.4420$$ $$P(X \ge 1) = 1 - (0.96)^{20} \approx 0.5580$$

For at least two, also take away $P(X = 1)$:

$$P(X = 1) = \binom{20}{1} (0.04)^1 (0.96)^{19} \approx 0.3683$$ $$P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \approx 1 - 0.4420 - 0.3683 = 0.1897$$

The expected number of defective bulbs is $np = 20(0.04) = 0.8$.

Common mistakes

Using the binomial formula when the trials are dependent. Drawing without replacement from a small group changes $p$ each time, so it isn’t binomial. Check all four conditions first.

Forgetting the combination. $p^k(1 - p)^{n - k}$ is the probability of one order. Multiply by $\binom{n}{k}$ to count all the orders.

Mixing up k and n − k in the exponents. The power of $p$ is the number of successes; the power of $1 - p$ is the number of failures. The two exponents always add to $n$.

Adding many terms for “at least”. $P(X \ge 1)$ for $20$ trials would need $20$ terms. Use $1 - P(X = 0)$ instead.

Getting “at most” and “at least” backwards. “At most $2$” means $X \le 2$ ($0$, $1$, or $2$). “At least $2$” means $X \ge 2$. Also, $P(X \gt 2)$ is $1 - P(X \le 2)$, not $1 - P(X \lt 2)$.

Practice

1. (Warm-up) Is $X$ binomial? Explain briefly.

• (a) $X$ is the number of heads in $8$ coin flips.
• (b) $X$ is the number of red marbles when $4$ are drawn without replacement from a bag of $5$ red and $5$ blue.
• (c) $X$ is the number of rolls of a die until you get a $1$.

Solution

(a) Yes: $n = 8$ independent flips, $p = 0.5$ each time.

(b) No: without replacement, the trials are dependent ($p$ changes after each draw).

(c) No: the number of trials isn’t fixed.

2. (Warm-up) A fair coin is flipped $6$ times. Find the probability of exactly $3$ heads.

Solution

$$P(X = 3) = \binom{6}{3} (0.5)^3 (0.5)^3 = \frac{20}{64} = \frac{5}{16} = 0.3125$$

3. (Warm-up) A binomial experiment has $n = 40$ and $p = 0.25$. Find $E(X)$.

Solution

$E(X) = np = 40(0.25) = 10$

4. (Core) A fair die is rolled $8$ times. Find the probability of exactly two $6$s.

Solution

Binomial with $n = 8$, $p = \tfrac{1}{6}$, $k = 2$:

$$P(X = 2) = \binom{8}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^6 = 28 \times \frac{1}{36} \times \frac{15\,625}{46\,656} \approx 0.2605$$

5. (Core) Make the probability distribution table for a binomial experiment with $n = 4$ and $p = 0.2$, and find $E(X)$. Describe the shape of its histogram.

Solution

Using $P(X = k) = \binom{4}{k}(0.2)^k(0.8)^{4 - k}$:

$k$	$0$	$1$	$2$	$3$	$4$
$P(X = k)$	$0.4096$	$0.4096$	$0.1536$	$0.0256$	$0.0016$

Check: the values add to $1$. ✓ $E(X) = 4(0.2) = 0.8$.

The histogram is strongly skewed right: two tall bars at $0$ and $1$, then a quick drop to almost nothing at $4$.

6. (Core) A student guesses on all $10$ questions of a multiple-choice quiz. Each question has $4$ choices.

• (a) Find the probability of getting at least one right.
• (b) Find the probability of getting at least $6$ right (a pass).
• (c) How many does the student expect to get right?

Solution

Binomial with $n = 10$, $p = 0.25$.

(a) $P(X \ge 1) = 1 - (0.75)^{10} \approx 1 - 0.0563 = 0.9437$.

(b) $P(X \ge 6) = P(6) + P(7) + P(8) + P(9) + P(10) \approx 0.0162 + 0.0031 + 0.0004 + 0.0000 + 0.0000 \approx 0.0197$. Guessing is not a good plan for passing.

(c) $E(X) = 10(0.25) = 2.5$ questions.

7. (Core) At a large school, $60\%$ of students walk or bike to school. Seven students are chosen at random. Find the probability that at most $2$ of them walk or bike.

Solution

Binomial with $n = 7$, $p = 0.6$. “At most $2$” means $X = 0$, $1$, or $2$.

$$\begin{aligned} P(X = 0) &= (0.4)^7 \approx 0.0016 \\ P(X = 1) &= \binom{7}{1}(0.6)(0.4)^6 \approx 0.0172 \\ P(X = 2) &= \binom{7}{2}(0.6)^2(0.4)^5 \approx 0.0774 \end{aligned}$$$$P(X \le 2) \approx 0.0963$$

(Adding the unrounded values gives $0.0963$; adding the three rounded values above gives $0.0962$. Keep full values on your calculator until the end.)

8. (Challenge) How many times must you roll a fair die so that the probability of getting at least one $6$ is more than $0.9$?

Solution

For $n$ rolls, $P(\text{at least one } 6) = 1 - \left(\tfrac{5}{6}\right)^n$. You need $\left(\tfrac{5}{6}\right)^n \lt 0.1$.

Try values: $\left(\tfrac{5}{6}\right)^{12} \approx 0.1122$ (too big) and $\left(\tfrac{5}{6}\right)^{13} \approx 0.0935$ (small enough).

So you need at least $13$ rolls. Then $P(\text{at least one } 6) \approx 0.9065$.

9. (Challenge) A binomial random variable has $n = 24$ and $E(X) = 6$. Find $p$, then find $P(X = 6)$. Is $6$ likely to be the single most probable value? Explain.

Solution

From $np = 6$: $p = \tfrac{6}{24} = 0.25$.

$$P(X = 6) = \binom{24}{6}(0.25)^6(0.75)^{18} = 134\,596 \times (0.25)^6 \times (0.75)^{18} \approx 0.1853$$

Yes. The tallest bar of a binomial histogram is at (or right next to) $np$, and here $np = 6$ is a whole number, so $6$ is the most probable value. Even so, its probability is only about $0.19$: the other values together are much more likely.