Capital $X$ names the random variable; lower-case $x$ is a particular value it takes. The distribution can be given as a table or as a probability function.
Two fair coins are tossed. $X$ is the number of heads. Find the probability distribution of $X$.
| $x$ | $0$ | $1$ | $2$ |
|---|---|---|---|
| $\mathrm{P}(X = x)$ | $\tfrac14$ | $\tfrac12$ | $\tfrac14$ |
$\mathrm{P}(X = x) = k(x + 1)$ for $x = 0, 1, 2, 3$, and $0$ otherwise. Find $k$ and $\mathrm{P}(X \geqslant 2)$.
e.g. the score on a fair six-sided die: $\mathrm{P}(X = x) = \tfrac16$ for $x = 1, 2, \ldots, 6$
Many situations count the number of "successes" in a fixed number of repeated trials: heads in $10$ tosses, defective items in a batch of $20$, seeds that germinate out of $12$.
② Each trial has two outcomes: success or failure
③ The trials are independent
④ The probability of success, $p$, is the same for every trial
Then $X$ = the number of successes
$X \sim \mathrm{B}(10, 0.3)$. Find $\mathrm{P}(X = 3)$.
Most questions ask for a range of values. Your calculator's binomial cumulative distribution function gives $\mathrm{P}(X \leqslant r)$ directly. (Some courses also provide printed tables of $\mathrm{P}(X \leqslant r)$.) Everything else is found from it.
| Wanted | Rewrite as |
|---|---|
| $\mathrm{P}(X < r)$ | $\mathrm{P}(X \leqslant r - 1)$ |
| $\mathrm{P}(X \geqslant r)$ | $1 - \mathrm{P}(X \leqslant r - 1)$ |
| $\mathrm{P}(X > r)$ | $1 - \mathrm{P}(X \leqslant r)$ |
| $\mathrm{P}(a \leqslant X \leqslant b)$ | $\mathrm{P}(X \leqslant b) - \mathrm{P}(X \leqslant a - 1)$ |
| $r$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
|---|---|---|---|---|---|---|---|---|
| $\mathrm{P}(X = r)$ | $0.0282$ | $0.1211$ | $0.2335$ | $0.2668$ | $0.2001$ | $0.1029$ | $0.0368$ | $0.0090$ |
| $\mathrm{P}(X \leqslant r)$ | $0.0282$ | $0.1493$ | $0.3828$ | $0.6496$ | $0.8497$ | $0.9527$ | $0.9894$ | $0.9984$ |
Values for $X \sim \mathrm{B}(10, 0.3)$, to 4 d.p.
$X \sim \mathrm{B}(10, 0.3)$. Find (a) $\mathrm{P}(X \leqslant 3)$, (b) $\mathrm{P}(X \geqslant 4)$, (c) $\mathrm{P}(2 \leqslant X \leqslant 5)$.
$Y \sim \mathrm{B}(20, 0.15)$. Find (a) $\mathrm{P}(Y < 4)$, (b) $\mathrm{P}(Y > 5)$.
Each ticket in a promotion has probability $0.2$ of winning a prize, independently. How many tickets must be bought for the probability of at least one prize to exceed $0.95$?
Seeds from a packet germinate with probability $0.85$. A gardener plants $10$ seeds in one tray and models the number $G$ that germinate as $\mathrm{B}(10, 0.85)$. (a) State two assumptions. (b) Find $\mathrm{P}(G \geqslant 9)$. (c) Comment on the assumptions.
Five cards are dealt from a standard pack. Explain why the number of hearts is not modelled exactly by $\mathrm{B}(5, 0.25)$.
Distribution
All values with their probabilities; total $1$.
Discrete uniform
$n$ equally likely values: each $\tfrac1n$.
Binomial conditions
Fixed $n$; two outcomes; independent; constant $p$.
Formula
$\mathrm{P}(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$
$\mathrm{P}(X < r)$
$= \mathrm{P}(X \leqslant r - 1)$
$\mathrm{P}(X \geqslant r)$
$= 1 - \mathrm{P}(X \leqslant r - 1)$
$\mathrm{P}(X > r)$
$= 1 - \mathrm{P}(X \leqslant r)$
At least one
$1 - (1-p)^n$
Context
Say what a trial is; question independence and constant $p$.
$\mathrm{P}(X = x) = kx^2$ for $x = 1, 2, 3$. Find $k$ and $\mathrm{P}(X = 3)$.
▶ Show solution
$k + 4k + 9k = 14k = 1$, so $k = \dfrac{1}{14}$.
$\mathrm{P}(X = 3) = \dfrac{9}{14}$.
$X$ has a discrete uniform distribution on $1, 2, \ldots, 8$. Find $\mathrm{P}(X > 5)$ and $\mathrm{P}(X \text{ is prime})$.
▶ Show solution
$\mathrm{P}(X > 5) = \mathrm{P}(6, 7, 8) = \dfrac38$.
Primes: $2, 3, 5, 7$, so $\mathrm{P} = \dfrac48 = \dfrac12$.
$X \sim \mathrm{B}(8, 0.4)$. Find $\mathrm{P}(X = 3)$.
▶ Show solution
$\dbinom{8}{3}(0.4)^3(0.6)^5 = 56 \times 0.064 \times 0.07776 = 0.2787$ (4 d.p.)
$X \sim \mathrm{B}(12, 0.25)$. Find $\mathrm{P}(X \leqslant 2)$ and $\mathrm{P}(X \geqslant 3)$.
▶ Show solution
$\mathrm{P}(X = 0) = 0.75^{12} = 0.0317$; $\mathrm{P}(X = 1) = 12(0.25)(0.75)^{11} = 0.1267$; $\mathrm{P}(X = 2) = 66(0.25)^2(0.75)^{10} = 0.2323$.
$\mathrm{P}(X \leqslant 2) = 0.3907$ (a calculator gives $0.39068$).
$\mathrm{P}(X \geqslant 3) = 1 - 0.3907 = 0.6093$.
$X \sim \mathrm{B}(15, 0.6)$. Find $\mathrm{P}(X > 10)$.
▶ Show solution
$\mathrm{P}(X > 10) = 1 - \mathrm{P}(X \leqslant 10) = 1 - 0.7827 = 0.2173$ (4 d.p.)
$X \sim \mathrm{B}(20, 0.35)$. Find $\mathrm{P}(5 \leqslant X < 9)$.
▶ Show solution
The values are $5, 6, 7, 8$.
$\mathrm{P}(X \leqslant 8) - \mathrm{P}(X \leqslant 4) = 0.7624 - 0.1182 = 0.6442$
A class of $30$ contains $12$ girls. Five students are chosen at random, one after another, to form a committee. Explain why the number of girls chosen does not exactly follow $\mathrm{B}(5, 0.4)$.
▶ Show solution
The students are chosen without replacement, so the probability of choosing a girl changes after each choice (e.g. $\tfrac{12}{30}$, then $\tfrac{11}{29}$ or $\tfrac{12}{29}$). The trials are not independent and $p$ is not constant. With such a small population the effect is noticeable.
$X \sim \mathrm{B}(n, 0.3)$. Find the smallest $n$ for which $\mathrm{P}(X = 0) < 0.01$.
▶ Show solution
$0.7^n < 0.01$, so $n > \dfrac{\ln 0.01}{\ln 0.7} = 12.9\ldots$
Check: $0.7^{12} = 0.0138$, $0.7^{13} = 0.0097$. So $n = 13$.
A spinner has $\mathrm{P}(\text{red}) = \tfrac16$. It is spun $6$ times. Find the probability of (a) exactly one red, (b) at least two reds.
▶ Show solution
$R \sim \mathrm{B}\left(6, \tfrac16\right)$.
(a) $6 \times \tfrac16 \times \left(\tfrac56\right)^5 = \left(\tfrac56\right)^5 = 0.4019$
(b) $\mathrm{P}(R = 0) = \left(\tfrac56\right)^6 = 0.3349$, so $\mathrm{P}(R \geqslant 2) = 1 - 0.3349 - 0.4019 = 0.2632$.
A basketball player scores each free throw with probability $0.7$. In a practice session she takes $15$ free throws.
(a) Define a suitable distribution for the number of scores, $S$, and state two assumptions needed. (b) Find $\mathrm{P}(S = 10)$. (c) Find $\mathrm{P}(S \geqslant 12)$. (d) She has $4$ such sessions. Find the probability that she scores at least $12$ in exactly $2$ of them. (e) Suggest why the assumptions might not hold.
▶ Show solution
(a) $S \sim \mathrm{B}(15, 0.7)$. Assumptions: the throws are independent; the probability of scoring is $0.7$ for every throw.
(b) $\mathrm{P}(S = 10) = \dbinom{15}{10}(0.7)^{10}(0.3)^5 = 3003 \times 0.0282475 \times 0.00243 = 0.2061$
(c) $\mathrm{P}(S \geqslant 12) = 1 - \mathrm{P}(S \leqslant 11) = 1 - 0.7031 = 0.2969$
(d) Let $T$ be the number of sessions with at least $12$: $T \sim \mathrm{B}(4, 0.29687\ldots)$.
$\mathrm{P}(T = 2) = \dbinom42 (0.29687)^2(0.70313)^2 = 6 \times 0.088131 \times 0.494395 = 0.2614$
(e) She may tire as the session goes on (so $p$ falls), or a miss may affect her confidence on the next throw (so throws are not independent).