Using the binomial model

What you will be able to do

• find "at least", "at most" and "between" probabilities, and the most likely value;
• find the number of trials needed to reach a target chance;
• use the normal approximation, with a continuity correction, for a large number of trials.

At least one success

For $n$ independent trials with success probability $p$, the chance of no success is $q^n$. So

How many trials make this at least a target $t$? Solve $1 - q^n \ge t$, which is $q^n \le 1 - t$. Take logarithms and remember that dividing by $\ln q$, a negative number, reverses the inequality:

Then round up to a whole number. For $p = 0.2$ and $t = 0.9$ we need $n \ge \dfrac{\ln 0.1}{\ln 0.8} \approx 10.3$, so $n = 11$. Ten trials give a chance of $0.8926$, which is just short, and eleven give $0.9141$.

At most, at least and between

Add the terms you need. For "at most $k$" add $P(X = 0)$ up to $P(X = k)$. For "at least $k$", use the complement when it is shorter:

Take $X \sim B(10, 0.2)$. Then $P(X \le 2) = 0.10737 + 0.26844 + 0.30199 \approx 0.6778$. So $P(X \ge 3) = 1 - 0.6778 = 0.3222$. A-Level and IB calculators give $P(X \le k)$ directly with the cumulative command, which AP calls binomcdf.

The most likely value

The tallest bar of a binomial distribution is at the integer part of $(n + 1)p$. If $(n + 1)p$ is itself a whole number, two neighbouring values tie. For $n = 10$ and $p = 0.35$, $(n + 1)p = 3.85$, so the most likely value is $3$.

When the trials are not independent

Drawing items from a small batch without replacement is not binomial. The exact model is called hypergeometric, and it has a slightly smaller variance. When the sample is a small part of a large population, the binomial is close. AP uses the rule that the sample should be less than $10\%$ of the population.

The first success

Let $T$ be the number of the trial on which the first success comes. For $T = k$, the first $k - 1$ trials fail and trial $k$ succeeds, so $P(T = k) = q^{k-1}p$. This is called the geometric distribution. Its mean is $\dfrac{1}{p}$: with $p = \dfrac{1}{6}$, you wait $6$ throws for a six on average.

The normal approximation

For large $n$, the bars of $B(n, p)$ form a bell shape. The normal distribution with the same mean and variance fits it well. A common rule is that both $np$ and $nq$ should be at least $5$ in A-Level texts, and at least $10$ in AP. Then

The binomial counts whole numbers, and the normal curve is continuous. Each bar at $k$ covers the interval from $k - 0.5$ to $k + 0.5$. The continuity correction takes account of this:

Here $\Phi$ is the distribution function of the standard normal distribution, from a table or a calculator. The Indian syllabus (NCERT, JEE Main) does not ask for this approximation. A-Level Statistics and AP Statistics do, and an IB student can use it to check a binomial answer.

For $n = 40$ and $p = 0.3$, $np = 12$ and $\sqrt{npq} \approx 2.898$. The exact value is $P(X \le 15) = 0.8849$. With the correction, $z = \dfrac{15.5 - 12}{2.898} \approx 1.208$, and $\Phi(1.208) \approx 0.8864$, which is close. Without the correction, $z = \dfrac{15 - 12}{2.898} \approx 1.035$, and $\Phi(1.035) \approx 0.8497$, which is not.

Practise with new numbers

Try it in Python

The program compares the exact binomial probability with the normal approximation, with and without the continuity correction.

Common mistakes

• Forgetting to round $n$ up. A target chance of at least $t$ needs the next whole number, not the nearest.
• Mixing "at least" and "more than". $P(X \ge 3)$ includes $3$, and $P(X > 3)$ does not.
• Skipping the continuity correction. Use $k + 0.5$ for $X \le k$, and $k - 0.5$ for $X \ge k$.
• Using the approximation for a small $n$. If $np$ or $nq$ is small, the bars are lopsided and the bell shape is a poor fit.

Remember these

• $P(\text{at least one}) = 1 - q^n$, and the number of trials needed is $n \ge \dfrac{\ln(1 - t)}{\ln q}$, rounded up.
• $P(X \ge k) = 1 - P(X \le k - 1)$.
• The most likely value of $B(n, p)$ is the integer part of $(n + 1)p$.
• For large $n$, $B(n, p) \approx N(np, npq)$, with a continuity correction of $\pm 0.5$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A fair die is thrown $6$ times. Find the probability of at least one six.

2. $X \sim B(8, 0.5)$. Find $P(X \le 2)$.

3. What is the smallest number of tosses of a fair coin for which the probability of at least one head is more than $0.999$?

4. Find the most likely value of $X \sim B(10, 0.35)$.

5. $X \sim B(64, 0.5)$. To approximate $P(X \le 28)$ with a normal distribution and a continuity correction, what is the $z$ value?

References

• OpenStax, Introductory Statistics, Chapter 4, sections on the binomial and geometric distributions, and Chapter 6 on the normal distribution.
• NCERT, Mathematics Part II for Class XII, Chapter 13, section on the binomial distribution.