PROBABILITY · HIGHER LEVEL
Bernoulli Trials & the Binomial Distribution
From coin tosses to a one-line formula · Tap NEXT to begin
Section 1 of 7
From Coin Tosses to a Formula
Before any new formula — let's see why we need it. Probability you already know how to do step-by-step. The Bernoulli formula is just a shortcut for a very specific shape of question. Once we see the shape, the formula writes itself.
(i) A coin is tossed 3 times — prob of 2 tails?
Don't reach for a formula yet. Just list out what we need:
We need T and T and H in some order.
One specific arrangement — say TTH — has probability:
$\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \,=\, \dfrac{1}{8}$
But TTH, THT and HTT all give 2 tails. How many such arrangements? Permutations of $\{T,T,H\}$:
$\dfrac{3!}{2!\,1!} \,=\, 3$ (2 Ts and 1 H)
$P(\text{2 tails}) \,=\, \dfrac{1}{8} \times 3 \,=\, \dfrac{3}{8}$
(ii) The hidden binomial coefficient
That number $3$ — the count of arrangements — is also $\binom{3}{2} \,=\, {}^{3}C_2$. So our calculation is really:
$P(\text{2 tails}) \,=\, {}^{3}C_2 \,\cdot\, \left(\dfrac{1}{2}\right)^{2} \,\cdot\, \left(\dfrac{1}{2}\right)^{1}$
Why does $\binom{3}{2}$ show up? Look at Pascal's triangle — row $n = 3$ is exactly $1,\,3,\,3,\,1$:
These four numbers are the coefficients in $(x+y)^3 \,=\, x^3 + 3x^2y + 3xy^2 + y^3$. They count how many arrangements there are of $r$ of one thing and $n-r$ of the other. That is exactly the count we need.
(iii) A coin is tossed 5 times — prob of 2 tails?
Same shape, bigger numbers. One arrangement, say TTHHH, has probability:
$\dfrac{1}{2} \,\cdot\, \dfrac{1}{2} \,\cdot\, \dfrac{1}{2} \,\cdot\, \dfrac{1}{2} \,\cdot\, \dfrac{1}{2} \,=\, \dfrac{1}{32}$
Number of arrangements with 2 Ts and 3 Hs:
$\dfrac{5!}{2!\,3!} \,=\, 10 \,=\, {}^{5}C_2$
$P(\text{2 tails}) \,=\, {}^{5}C_2 \,\cdot\, \left(\dfrac{1}{2}\right)^{2} \,\cdot\, \left(\dfrac{1}{2}\right)^{3} \,=\, \dfrac{10}{32} \,=\, \dfrac{5}{16}$
Notice the pattern: ${}^{n}C_r \cdot (\text{success})^{r} \cdot (\text{failure})^{n-r}$. Every Bernoulli-trials question collapses to that one line. Next we just need to know when we're allowed to use it.
You try
A coin is tossed 4 times. Find the probability of getting exactly 3 heads.
List: one arrangement is HHHT. There are $\binom{4}{3}$ such arrangements.
One arrangement (e.g. HHHT) has probability $\left(\tfrac{1}{2}\right)^{4} \,=\, \dfrac{1}{16}$.
Number of such arrangements: $\binom{4}{3} \,=\, 4$.
$P(\text{3 heads}) \,=\, 4 \,\cdot\, \dfrac{1}{16} \,=\, \dfrac{4}{16} \,=\, \dfrac{1}{4}$.
$\dfrac{1}{4}$
$\dfrac{1}{4}$
You try
A coin is tossed 6 times. Find the probability of getting exactly 4 tails.
Same idea: $\binom{6}{4}$ arrangements, each with probability $(1/2)^6$.
One arrangement (e.g. TTTTHH) has probability $\left(\tfrac{1}{2}\right)^{6} \,=\, \dfrac{1}{64}$.
Number of arrangements with 4 Ts and 2 Hs: $\binom{6}{4} \,=\, 15$.
$P(\text{4 tails}) \,=\, 15 \,\cdot\, \dfrac{1}{64} \,=\, \dfrac{15}{64}$.
$\dfrac{15}{64}$
$\dfrac{15}{64}$
Section 2 of 7
The Bernoulli Conditions
That coin-toss formula works in a very specific situation — when every trial behaves the same way. Probability questions where this happens are called Bernoulli trials (or binomial — same thing, two names).
To use the formula, all four of these must be true:
Bernoulli trials — the 4 conditions (must learn)
(i)There are only 2 possible outcomes on each trial. Call them success and failure: $p \,=\, P(\text{success})$, $q \,=\, P(\text{failure}) \,=\, 1 - p$.
(ii)There are a fixed number of trials $\,n$.
(iii)The trials are independent. (one trial doesn't affect another)
(iv)The probability of success on each trial is a constant $\,p$. (it doesn't change from trial to trial)
(i) Two outcomes per trial — what counts?
You'd expect coin tosses (H or T) to qualify. But the "2 outcomes" idea is broader than that — you just need to split the world into the thing you want versus everything else.
| Situation | Success ($p$) | Failure ($q$) |
|---|---|---|
| Toss a coin | Heads | Tails |
| Take a free throw | Score | Miss |
| Roll a die — get a 6? | Roll a 6 ($p\!=\!\tfrac{1}{6}$) | Anything else ($q\!=\!\tfrac{5}{6}$) |
| Pick a card — get a heart? | Heart ($p\!=\!\tfrac{1}{4}$) | Not a heart ($q\!=\!\tfrac{3}{4}$) |
| Answer a phone call | Answer | Miss |
"Two outcomes" doesn't mean the experiment only has two physical outcomes — it means we classify the outcome into "the thing I want" vs "anything else". Rolling a die has six outcomes, but if the question is "did you roll a 6?" then there are two outcomes for our purposes.
(ii) When does Bernoulli NOT apply?
Watch out — these break the conditions
✗Drawing without replacement. If you draw cards from a deck and don’t put them back, then $p$ changes after each draw. Condition (iv) fails.
✗Sequence matters. If the question pins down a specific order (e.g. “first and last only”), then ${}^{n}C_r$ doesn’t apply — see Section 5.
✗Number of trials isn’t fixed. “How many tosses until I get heads?” is not Bernoulli — $n$ isn’t set.
You try
I roll a fair die 10 times and count how many sixes I get. Is this a Bernoulli situation? Briefly check each condition.
Walk through the four conditions. Each trial: roll a die.
(i) Two outcomes? Yes — "six" (success, $p = \tfrac{1}{6}$) vs "not a six" (failure, $q = \tfrac{5}{6}$). ✓
(ii) Fixed $n$? Yes, $n = 10$. ✓
(iii) Independent? Yes — one roll has no effect on the next. ✓
(iv) Constant $p$? Yes, $p = \tfrac{1}{6}$ every time. ✓
All four conditions hold — this is Bernoulli.
Yes — all four conditions are satisfied.
You try
From a deck of 52 cards, I draw 5 cards without replacement and count how many are hearts. Is this Bernoulli?
Check condition (iv) carefully — does $p$ stay the same after each draw?
(i) Two outcomes? Yes — heart (success) vs not. ✓
(ii) Fixed $n = 5$? Yes. ✓
(iii) Independent? No — what I draw on card 1 changes what's in the deck for card 2.
(iv) Constant $p$? No — $p$ starts at $\tfrac{13}{52}$ but if I drew a heart it drops to $\tfrac{12}{51}$, and so on.
Not Bernoulli — without replacement breaks (iii) and (iv).
No — drawing without replacement breaks the constant-$p$ condition.
Section 3 of 7
The Binomial Formula
Once a question passes the four Bernoulli tests, the probability of getting exactly $r$ successes in $n$ trials is given by one line of maths:
The Binomial Formula — must learn
$P(r) \,=\, \binom{n}{r} \,p^{\,r} \,q^{\,n-r}$
•$n$ = total number of trials (fixed)
•$r$ = number of required successes
•$p$ = probability of success on each trial
•$q$ = probability of failure = $1 - p$
•$\binom{n}{r} \,=\, {}^{n}C_r$ = number of ways to arrange $r$ successes among $n$ trials
The formula does three jobs at once. The $\binom{n}{r}$ counts the arrangements, $p^{\,r}$ accounts for the $r$ successes, $q^{\,n-r}$ accounts for the $n-r$ failures.
(i) The four-step routine for every Bernoulli question
How to answer a Bernoulli question
1.Identify the trial. What counts as success? What is $p$?
2.Write down $n$, $r$, $p$, $q$ (the four numbers — always set these out at the top of your answer).
3.Substitute into $P(r) \,=\, \binom{n}{r} p^{\,r} q^{\,n-r}$.
4.Evaluate — usually to 2 decimal places.
(ii) Calculator — finding ⁿCᵣ on a Casio
Casio fx-83 / 85 / 991
•Type the value of $n$. SHIFT → the button labelled nCr (above the ÷ key on most models). Type $r$. Press =.
•Example: $5\;\text{SHIFT}\;\text{nCr}\;2\;=\;10$.
•On older models the button is labelled C.
You try
For each of the following, write down $n$, $r$, $p$ and $q$. Don't solve — just set the numbers out.
(a) A coin is tossed 8 times. Prob of exactly 3 tails?
(b) Prob of scoring is $0.6$. I take 10 shots. Prob of exactly 7 scores?
(c) I roll a die 12 times. Prob of getting exactly 4 sixes?
(a) A coin is tossed 8 times. Prob of exactly 3 tails?
(b) Prob of scoring is $0.6$. I take 10 shots. Prob of exactly 7 scores?
(c) I roll a die 12 times. Prob of getting exactly 4 sixes?
Success and failure must add to 1. Then identify $n$ (total trials) and $r$ (required successes).
(a) $n = 8, \quad r = 3, \quad p = \tfrac{1}{2}, \quad q = \tfrac{1}{2}$.
(b) $n = 10, \quad r = 7, \quad p = 0.6, \quad q = 0.4$.
(c) $n = 12, \quad r = 4, \quad p = \tfrac{1}{6}, \quad q = \tfrac{5}{6}$.
(a) $n\!=\!8,r\!=\!3,p\!=\!q\!=\!\tfrac{1}{2}$ · (b) $n\!=\!10,r\!=\!7,p\!=\!0.6,q\!=\!0.4$ · (c) $n\!=\!12,r\!=\!4,p\!=\!\tfrac{1}{6},q\!=\!\tfrac{5}{6}$
Section 4 of 7
Worked Example — 5 Tosses, 2 Tails
Same question from Section 1 — but now we use the formula instead of building it up from scratch.
(i) Set out the four numbers
$n \,=\, 5 \qquad r \,=\, 2 \qquad p \,=\, \dfrac{1}{2} \qquad q \,=\, \dfrac{1}{2}$
(ii) Substitute into the formula
$P(r) \,=\, \binom{n}{r} p^{\,r} q^{\,n-r}$
$P(2) \,=\, \binom{5}{2} \,\left(\dfrac{1}{2}\right)^{2} \,\left(\dfrac{1}{2}\right)^{3}$
(iii) Evaluate
$\binom{5}{2} \,=\, 10$ (SHIFT nCr on calculator)
$\left(\dfrac{1}{2}\right)^{2} \,=\, \dfrac{1}{4} \qquad \left(\dfrac{1}{2}\right)^{3} \,=\, \dfrac{1}{8}$
$P(2) \,=\, 10 \,\cdot\, \dfrac{1}{4} \,\cdot\, \dfrac{1}{8} \,=\, \dfrac{10}{32}$
$P(2) \,=\, \dfrac{5}{16}$
Same answer as Section 1, much less working. That is what the formula is for.
You try
A coin is tossed 7 times. Find the probability of exactly 4 heads.
$n=7, r=4, p=q=\tfrac{1}{2}$. Plug straight into the formula.
$n = 7 \qquad r = 4 \qquad p = q = \dfrac{1}{2}$
$P(4) \,=\, \binom{7}{4} \,\left(\dfrac{1}{2}\right)^{4} \,\left(\dfrac{1}{2}\right)^{3}$
$\hphantom{P(4)} \,=\, 35 \,\cdot\, \dfrac{1}{16} \,\cdot\, \dfrac{1}{8} \,=\, \dfrac{35}{128}$.
$P(4) \,=\, \dfrac{35}{128} \,\approx\, 0.27$
$\dfrac{35}{128} \approx 0.27$
You try
A spinner has $P(\text{red}) = 0.4$. The spinner is spun 6 times. Find the probability of getting exactly 2 reds.
$n=6, r=2, p=0.4, q=0.6$.
$n = 6 \qquad r = 2 \qquad p = 0.4 \qquad q = 0.6$
$P(2) \,=\, \binom{6}{2} \,(0.4)^{2} \,(0.6)^{4}$
$\hphantom{P(2)} \,=\, 15 \,\cdot\, 0.16 \,\cdot\, 0.1296$
$P(2) \,\approx\, 0.31$
$\approx 0.31$
Section 5 of 7
The Four Question Types: Free Throws
Carl's favourite Bernoulli example — and it shows up in every exam paper. Each variation looks similar but uses the formula slightly differently. Learn the four flavours and you can read any exam question.
Setup — used in all four parts
The probability I score a free throw is $\dfrac{3}{7}$. I take 5 free throws.
→$p \,=\, \dfrac{3}{7} \qquad q \,=\, \dfrac{4}{7} \qquad n \,=\, 5$
(i) Score exactly 3 — the standard one
Question. What is the probability I score exactly 3 of the 5?
$n \,=\, 5 \qquad r \,=\, 3 \qquad p \,=\, \dfrac{3}{7} \qquad q \,=\, \dfrac{4}{7}$
$P(3) \,=\, \binom{5}{3} \,\left(\dfrac{3}{7}\right)^{3} \,\left(\dfrac{4}{7}\right)^{2}$
$\hphantom{P(3)} \,=\, 10 \,\cdot\, \dfrac{27}{343} \,\cdot\, \dfrac{16}{49}$
$\hphantom{P(3)} \,=\, 0.257$
$\approx 0.26$
(ii) Score the first and last only — order is fixed
Question. What is the probability I score only the first and last free throws?
Hold on — read this carefully. "Score only the first and last" means the sequence is locked:
S M M M S
Only one arrangement gives this outcome — so there's nothing to count. Don't reach for $\binom{n}{r}$. Just multiply the probabilities in order:
$P \,=\, \dfrac{3}{7} \,\cdot\, \dfrac{4}{7} \,\cdot\, \dfrac{4}{7} \,\cdot\, \dfrac{4}{7} \,\cdot\, \dfrac{3}{7}$
$\hphantom{P} \,=\, \left(\dfrac{3}{7}\right)^{2} \,\left(\dfrac{4}{7}\right)^{3}$
$\hphantom{P} \,\approx\, 0.034$
Bernoulli takes care of order
When a question specifies a particular sequence (“first and last”, “the 2nd and 4th”), the order is fixed — only 1 arrangement counts. Do NOT use $\binom{n}{r}$.
When a question says “any” or “exactly $r$”, the order is free — use the full Bernoulli formula with $\binom{n}{r}$.
(iii) Score any 2 of the 5 — standard Bernoulli
Question. Find the probability I score any 2 of the 5.
"Any 2" — so order is free. Now $\binom{n}{r}$ is back in:
$n \,=\, 5 \qquad r \,=\, 2 \qquad p \,=\, \dfrac{3}{7} \qquad q \,=\, \dfrac{4}{7}$
$P(2) \,=\, \binom{5}{2} \,\left(\dfrac{3}{7}\right)^{2} \,\left(\dfrac{4}{7}\right)^{3}$
$\hphantom{P(2)} \,\approx\, 0.34$
(iv) Score for the 3rd time on the 5th free — the split trick
Question. What is the probability I score for the 3rd time on the 5th free throw?
Pause — read this slowly. "Score for the 3rd time on the 5th free" means:
Decoding the question
•After all 5 throws, I have scored exactly 3.
•The 5th throw was the one that pushed me to 3 — so the 5th must be a score.
•Which means in the first 4 throws I scored exactly 2 (in any order).
So the answer splits in two:
| First 4 throws have any 2 scores: | $\binom{4}{2} \,\left(\dfrac{3}{7}\right)^{2} \left(\dfrac{4}{7}\right)^{2}$ |
| 5th throw is a score: | $\dfrac{3}{7}$ |
$P \,=\, \binom{4}{2} \,\left(\dfrac{3}{7}\right)^{2} \left(\dfrac{4}{7}\right)^{2} \,\cdot\, \dfrac{3}{7}$
$\hphantom{P} \,=\, 6 \,\cdot\, \dfrac{9}{49} \,\cdot\, \dfrac{16}{49} \,\cdot\, \dfrac{3}{7}$
$\hphantom{P} \,\approx\, 0.15$
The trick: the LAST throw is forced to be a success, so peel it off. Apply Bernoulli to the first $n-1$ trials with $r-1$ successes, then multiply by $p$ for the final trial.
The four flavours — at a glance
(i)Exactly $r$ successes (any order). Use $\binom{n}{r} p^{\,r} q^{\,n-r}$.
(ii)A particular sequence (e.g. “first and last only”). Multiply the probabilities in order — no $\binom{n}{r}$.
(iii)Any $r$ — same as (i).
(iv)$r$-th success on the $n$-th trial. Apply Bernoulli to the first $n-1$ trials (for $r-1$ successes), then multiply by $p$.
Section 6 of 7
Phone Calls — All Three Types
Same four flavours, new numbers. Use this example to test yourself before moving on.
Setup — used in all three parts
The probability I answer a phone call is $\dfrac{2}{9}$. In 7 calls find the probability that…
→$p \,=\, \dfrac{2}{9} \qquad q \,=\, \dfrac{7}{9} \qquad n \,=\, 7$
You try
(i) I answer only the 1st and last calls. Find this probability.
Specific sequence — order is fixed. Don't use $\binom{n}{r}$.
Sequence is locked: A M M M M M A
$P \,=\, \dfrac{2}{9} \,\cdot\, \dfrac{7}{9} \,\cdot\, \dfrac{7}{9} \,\cdot\, \dfrac{7}{9} \,\cdot\, \dfrac{7}{9} \,\cdot\, \dfrac{7}{9} \,\cdot\, \dfrac{2}{9}$
$\hphantom{P} \,=\, \left(\dfrac{2}{9}\right)^{2} \left(\dfrac{7}{9}\right)^{5}$
$P \,\approx\, 0.014$
$\approx 0.014$
You try
(ii) I answer any 2 of the 7 calls. Find this probability.
"Any 2" — standard Bernoulli. $n=7,r=2,p=\tfrac{2}{9},q=\tfrac{7}{9}$.
$n = 7 \qquad r = 2 \qquad p = \dfrac{2}{9} \qquad q = \dfrac{7}{9}$
$P(2) \,=\, \binom{7}{2} \,\left(\dfrac{2}{9}\right)^{2} \,\left(\dfrac{7}{9}\right)^{5}$
$\hphantom{P(2)} \,=\, 21 \,\cdot\, \dfrac{4}{81} \,\cdot\, \dfrac{16807}{59049}$
$P(2) \,\approx\, 0.30$
$\approx 0.30$
You try
(iii) I answer for the 3rd time on the 7th call. Find this probability.
Split: 2 of the first 6 are answered (Bernoulli), then the 7th must be answered (× $p$).
The 7th call is forced — it has to be an answer.
So in the first 6 calls I answer exactly 2 in any order.
First 6: $n = 6, \, r = 2, \, p = \dfrac{2}{9}, \, q = \dfrac{7}{9}$
$P \,=\, \binom{6}{2} \,\left(\dfrac{2}{9}\right)^{2} \left(\dfrac{7}{9}\right)^{4} \,\cdot\, \dfrac{2}{9}$
$\hphantom{P} \,=\, 15 \,\cdot\, \dfrac{4}{81} \,\cdot\, \dfrac{2401}{6561} \,\cdot\, \dfrac{2}{9}$
$P \,\approx\, 0.06$
$\approx 0.06$
Section 7 of 7
Mixed Practice
Each of these is one of the four flavours. Read carefully — half the work is spotting which one.
You try
A football team has a $0.7$ probability of winning any given match. In a 10-match season, find the probability the team wins exactly 8 matches.
Standard Bernoulli — flavour (i).
$n = 10 \qquad r = 8 \qquad p = 0.7 \qquad q = 0.3$
$P(8) \,=\, \binom{10}{8} \,(0.7)^{8} \,(0.3)^{2}$
$\hphantom{P(8)} \,=\, 45 \,\cdot\, 0.0576 \,\cdot\, 0.09$
$P(8) \,\approx\, 0.23$
$\approx 0.23$
You try
A drug works with probability $0.8$ in any patient. In 12 patients, find the probability the drug works in exactly 10.
$n=12, r=10, p=0.8, q=0.2$ — flavour (i).
$n = 12 \qquad r = 10 \qquad p = 0.8 \qquad q = 0.2$
$P(10) \,=\, \binom{12}{10} \,(0.8)^{10} \,(0.2)^{2}$
$\hphantom{P(10)} \,=\, 66 \,\cdot\, 0.1074 \,\cdot\, 0.04$
$P(10) \,\approx\, 0.28$
$\approx 0.28$
You try
$P(\text{hit the target}) = \dfrac{1}{4}$ on each shot. I take 6 shots. Find the probability that I hit the target on the 2nd and 5th shots only.
Specific sequence — flavour (ii). No $\binom{n}{r}$. Miss, Hit, Miss, Miss, Hit, Miss.
Sequence is locked: M H M M H M
$P \,=\, \dfrac{3}{4} \,\cdot\, \dfrac{1}{4} \,\cdot\, \dfrac{3}{4} \,\cdot\, \dfrac{3}{4} \,\cdot\, \dfrac{1}{4} \,\cdot\, \dfrac{3}{4}$
$\hphantom{P} \,=\, \left(\dfrac{1}{4}\right)^{2} \left(\dfrac{3}{4}\right)^{4} \,=\, \dfrac{81}{4096}$
$P \,\approx\, 0.020$
$\dfrac{81}{4096} \,\approx\, 0.020$
You try
A basketball player has a free-throw success rate of $0.8$. He takes 6 throws. Find the probability he scores for the 4th time on the 6th throw.
Split: any 3 of the first 5 are scores (Bernoulli), then × $p$ for the 6th.
$p = 0.8, \, q = 0.2$. The 6th throw must be a score.
First 5 throws: $n = 5, \, r = 3$
$P \,=\, \binom{5}{3} \,(0.8)^{3} \,(0.2)^{2} \,\cdot\, 0.8$
$\hphantom{P} \,=\, 10 \,\cdot\, 0.512 \,\cdot\, 0.04 \,\cdot\, 0.8$
$P \,\approx\, 0.16$
$\approx 0.16$
You try
A die is rolled 8 times. Find the probability of getting exactly 3 sixes.
$p = \tfrac{1}{6}, q = \tfrac{5}{6}$. Flavour (i).
$n = 8 \qquad r = 3 \qquad p = \dfrac{1}{6} \qquad q = \dfrac{5}{6}$
$P(3) \,=\, \binom{8}{3} \,\left(\dfrac{1}{6}\right)^{3} \,\left(\dfrac{5}{6}\right)^{5}$
$\hphantom{P(3)} \,=\, 56 \,\cdot\, \dfrac{1}{216} \,\cdot\, \dfrac{3125}{7776}$
$P(3) \,\approx\, 0.10$
$\approx 0.10$
You try
An archer hits the bullseye with probability $0.6$. Find the probability she hits the bullseye for the 2nd time on her 4th shot.
Split — any 1 of the first 3 is a hit, then × $p$ for the 4th shot.
$p = 0.6, \, q = 0.4$. The 4th shot must be a hit.
First 3 shots: $n = 3, \, r = 1$
$P \,=\, \binom{3}{1} \,(0.6)^{1} \,(0.4)^{2} \,\cdot\, 0.6$
$\hphantom{P} \,=\, 3 \,\cdot\, 0.6 \,\cdot\, 0.16 \,\cdot\, 0.6$
$P \,\approx\, 0.17$
$\approx 0.17$
You're done.
Bernoulli Trials & the Binomial Distribution · Higher Level · Mathslive.ie