4.8 The binomial distributionIB Maths: Analysis and Approaches SL: Revision notes

A discrete random variable $X$ follows a binomial distribution, written $X\sim B(n,p)$, when it counts the number of successes in a fixed number of trials and all four conditions hold:

• there is a fixed number of trials, $n$;
• each trial has exactly two outcomes, success or failure;
• the probability of success, $p$, is the same on every trial;
• the trials are independent.

For example, the number of correct answers when 12 four-option questions are guessed at random is $B(12,0.25)$. Taking 5 cards from a pack of 52 without replacement and counting hearts is not binomial, because $p$ changes after each card. In context questions you may be asked to criticise the model: people travelling in groups, or machines that fail in batches, break independence.

Binomial probabilities are found with technology. On a GDC:

• binomial pdf gives $P(X=x)$;
• binomial cdf gives $P(X\le x)$.

The formula $P(X=r)=\binom{n}{r}p^{r}(1-p)^{n-r}$ is in the formula booklet, but in an exam you use the calculator. Translate the words carefully, remembering that $X$ takes whole-number values only:

• 'at most 5': $P(X\le5)$
• 'fewer than 5': $P(X\le4)$
• 'at least 5': $P(X\ge5)=1-P(X\le4)$
• 'more than 5': $P(X\ge6)=1-P(X\le5)$
• 'between 3 and 7 inclusive': $P(X\le7)-P(X\le2)$

For $X\sim B(n,p)$: $\mathrm{E}(X)=np, \qquad \mathrm{Var}(X)=np(1-p).$ The standard deviation is $\sqrt{np(1-p)}$. Both results are in the formula booklet, and you do not need to prove them.

The mean is the long-run average number of successes, so it does not have to be a whole number: 186 ticket holders turning up with probability 0.95 gives a mean of $176.7$. The most likely value (the mode) is the value of $x$ with the largest $P(X=x)$. You can find it from a table of values on the GDC. It is usually close to the mean but need not be equal to it.

Some questions fix a probability and ask for the least or greatest value of $n$.

Using the complement. 'At least one miss' is the complement of 'no misses', so $P(\text{at least one miss})=1-p^{n}$, where $p$ is the probability of a success. Solving $1-(0.78)^n>0.99$ gives $(0.78)^n<0.01$, so $n>\frac{\ln0.01}{\ln0.78}=18.5$ and the least $n$ is 19. The inequality reverses when you divide by $\ln0.78$, because it is negative.

Trial and improvement. When the condition involves a cumulative probability, such as $P(X>180)<0.05$ with $X\sim B(n,0.95)$, use a GDC table in $n$. Quote the values for the two consecutive values of $n$ on either side of the boundary. This is your justification.

A probability from one binomial model can be the $p$ of a second model. If a flight is overbooked with probability $0.0433$ on any day, and days are independent, then the number of overbooked days in a week is $Y\sim B(7,0.0433)$. So $P(Y\ge1)=1-(1-0.0433)^{7}=0.266.$ Keep full calculator accuracy for the first probability, and round only the final answer.