2. Each trial has just two outcomes: success or failure
3. The trials are independent
4. The probability of success, $p$, is the same every time
| Situation | Binomial? | Why |
|---|---|---|
| $20$ coin tosses, counting heads | Yes | $n=20$, $p=0.5$, independent |
| $5$ dice, counting sixes | Yes | $n=5$, $p=\tfrac16$ |
| $10$ bulbs tested, $4\%$ faulty | Yes | $n=10$, $p=0.04$ |
| $3$ cards dealt, counting aces | No | Without replacement — $p$ changes |
| Rolling until a six appears | No | $n$ is not fixed |
A biased coin has $P(\text{head}) = 0.3$. It is tossed $4$ times. Find $P(\text{exactly 2 heads})$.
A die is rolled $8$ times. Find the probability of exactly $3$ sixes.
$P(\text{at least } 2) = 1 - P(0) - P(1)$
| The words | Means | Best route |
|---|---|---|
| exactly $r$ | $X = r$ | One formula |
| at least $1$ | $X \geqslant 1$ | $1 - P(0)$ |
| at least $2$ | $X \geqslant 2$ | $1 - P(0) - P(1)$ |
| at most $2$ | $X \leqslant 2$ | $P(0)+P(1)+P(2)$ |
| fewer than $3$ | $X \leqslant 2$ | Same as above |
| more than $3$ | $X \geqslant 4$ | $1 - P(0) - P(1) - P(2) - P(3)$ |
Find the probability of obtaining at least two sixes when five dice are rolled.
$8\%$ of items from a production line are faulty. A sample of $6$ is taken. Find the probability that at most one is faulty.
A multiple-choice test has $40$ questions, each with $5$ options. A student guesses every answer. How many would you expect them to get right?
Conditions
Fixed $n$, two outcomes, independent, constant $p$.
Formula
$\binom{n}{r}p^r(1-p)^{n-r}$.
Why $\binom{n}{r}$
It counts the paths with $r$ successes.
At least $1$
$1 - P(0)$.
At least $2$
$1 - P(0) - P(1)$.
At most $2$
$P(0)+P(1)+P(2)$.
Fewer than $3$
Same as at most $2$.
Expected value
$np$.
Not binomial
Without replacement, or $n$ not fixed.
Accuracy
Keep full figures; round at the end.
A fair coin is tossed $5$ times. Find $P(\text{exactly 3 heads})$.
▶ Show solution
$\binom{5}{3}(0.5)^3(0.5)^2 = 10 \times (0.5)^5 = \dfrac{10}{32} = 0.3125$
A die is rolled $4$ times. Find $P(\text{exactly one six})$.
▶ Show solution
$\binom{4}{1}\left(\tfrac16\right)\left(\tfrac56\right)^3 = 4 \times \tfrac16 \times \dfrac{125}{216}$
$= \dfrac{500}{1296} = 0.386$ (3 s.f.)
$15\%$ of seeds fail to germinate. Ten are planted. Find $P(\text{all germinate})$.
▶ Show solution
$P(\text{one germinates}) = 0.85$
$P(\text{all ten}) = (0.85)^{10} = 0.1969 = 0.197$ (3 s.f.)
Using the data of Question 3, find the probability that at least one seed fails.
▶ Show solution
$P(\text{at least one fails}) = 1 - P(\text{none fails})$
$= 1 - 0.1969 = 0.803$ (3 s.f.)
A basketball player scores $70\%$ of free throws. She takes $6$. Find $P(\text{exactly 5 scored})$.
▶ Show solution
$\binom{6}{5}(0.7)^5(0.3)^1 = 6 \times 0.16807 \times 0.3$
$= 0.3025 = 0.303$ (3 s.f.)
A test has $25$ questions, each with $4$ options. How many would a guesser expect to get right?
▶ Show solution
$\text{E}(X) = np = 25 \times 0.25 = 6.25$ questions.
Six dice are rolled. Find the probability of at least one six.
▶ Show solution
$P(\text{no sixes}) = \left(\tfrac56\right)^6 = \dfrac{15625}{46656} = 0.33490$
$P(\text{at least one}) = 1 - 0.33490 = 0.665$ (3 s.f.)
$12\%$ of a large batch of components are faulty. Five are chosen. Find the probability that at most one is faulty.
▶ Show solution
$P(X=0) = (0.88)^5 = 0.52773$
$P(X=1) = \binom{5}{1}(0.12)(0.88)^4 = 5 \times 0.12 \times 0.59969 = 0.35981$
$P(X \leqslant 1) = 0.52773 + 0.35981 = 0.888$ (3 s.f.)
Explain why the binomial distribution should not be used to find the probability of drawing exactly $2$ aces when $5$ cards are dealt from a pack of $52$. Find the correct probability.
▶ Show solution
The cards are dealt without replacement, so the probability of an ace changes after each card. Conditions 3 and 4 fail — the trials are neither independent nor identically distributed.
Use combinations instead. Choose $2$ aces from $4$ and $3$ non-aces from $48$:
$P = \dfrac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}} = \dfrac{6 \times 17\,296}{2\,598\,960}$
$= \dfrac{103\,776}{2\,598\,960} = 0.0399$ (3 s.f.)
For comparison, the (incorrect) binomial answer with $p = \tfrac{4}{52}$ would be $\binom{5}{2}\left(\tfrac{1}{13}\right)^2\left(\tfrac{12}{13}\right)^3 = 0.0451$ — close, but wrong.
A quality inspector tests $10$ items from a large batch in which $5\%$ are known to be faulty. The batch is rejected if $2$ or more of the sample are faulty.
(a) Find the probability that the batch is accepted. (b) Find the expected number of faulty items in the sample. (c) The inspector wants the probability of accepting a $5\%$-faulty batch to be below $50\%$. Show that testing $10$ items cannot achieve this, and find the smallest sample size that does. (d) Comment on whether the "$2$ or more" rule is a good test.
▶ Show solution
(a) Accepted means $X \leqslant 1$, with $n=10$, $p=0.05$.
$P(X=0) = (0.95)^{10} = 0.598737$
$P(X=1) = \binom{10}{1}(0.05)(0.95)^9 = 10 \times 0.05 \times 0.630249 = 0.315125$
$P(\text{accepted}) = 0.598737 + 0.315125 = 0.9139 = 0.914$ (3 s.f.)
(b) $\text{E}(X) = 10 \times 0.05 = 0.5$ faulty items.
(c) From (a), with $n=10$ the acceptance probability is $0.914$, far above $0.5$. So $10$ items is not enough.
We need $P(X \leqslant 1) < 0.5$, i.e. $(0.95)^n + n(0.05)(0.95)^{n-1} < 0.5$.
Trying values:
$n = 30$: $(0.95)^{30} = 0.21464$; $30(0.05)(0.95)^{29} = 1.5 \times 0.22594 = 0.33891$; total $0.5536$ — still too high.
$n = 33$: $(0.95)^{33} = 0.18406$; $33(0.05)(0.95)^{32} = 1.65 \times 0.19375 = 0.31969$; total $0.5037$ — just too high.
$n = 34$: $(0.95)^{34} = 0.17485$; $34(0.05)(0.95)^{33} = 1.7 \times 0.18406 = 0.31290$; total $0.4878 < 0.5$ ✓
The smallest sample size is $n = 34$.
(d) The rule is a weak test as it stands. A batch with the full $5\%$ fault rate is accepted about $91\%$ of the time, so the inspection lets almost every bad batch through.
Part (c) shows the problem is the sample size, not the threshold: with only $10$ items you expect just $0.5$ faults, so seeing $2$ is genuinely unusual even when the batch is bad. To make the test discriminating, the inspector must either test many more items (at least $34$) or reject on a single fault — though that would then reject good batches too often.