Discrete random variables and the binomial distribution – IB Maths exam questions with mark scheme

1StandardCalculator6 marks

The random variable $X$ follows a binomial distribution with $n=12$ and $p=0.3$.

(a) Find $P(X=4)$. [2]

(b) Find $P(X\ge3)$. [2]

$P(X=4)=\binom{12}{4}(0.3)^4(0.7)^8$ (or GDC binomial pdf)M1
$0.231$ ($0.231139\ldots$)A1
$P(X\ge3)=1-P(X\le2)$ (or GDC binomial cdf with lower bound 3, upper bound 12)M1
$0.747$ ($0.747184\ldots$)A1
$E(X)=np=3.6$A1
$\mathrm{Var}(X)=np(1-p)=2.52$A1
Answer: (a) $0.231$ (b) $0.747$ (c) $E(X)=3.6$, $\mathrm{Var}(X)=2.52$
Examiner tip: $P(X\ge3)$ is $1-P(X\le2)$, not $1-P(X\le3)$; check the boundary value.

2StandardCalculator7 marks

The discrete random variable $X$ has probability distribution $P(X=x)=k(x+1)$ for $x=0,1,2,3$.

(a) Find the value of $k$ and write down the probability distribution as a table. [3]

(b) Find $E(X)$. [2]

$\sum P(X=x)=1$: $k+2k+3k+4k=1$M1
$k=0.1$A1
Table: $P(X=0,1,2,3)=0.1,\ 0.2,\ 0.3,\ 0.4$A1
$E(X)=\sum xP(X=x)=0(0.1)+1(0.2)+2(0.3)+3(0.4)$M1
$E(X)=2$A1
$E(X^2)=0.2+1.2+3.6=5$ and $\mathrm{Var}(X)=E(X^2)-[E(X)]^2$M1
$\mathrm{Var}(X)=5-4=1$A1
Answer: (a) $k=0.1$ (b) $2$ (c) $1$
Examiner tip: Check your probabilities sum to 1 before moving on, and use $E(X^2)-[E(X)]^2$ rather than forgetting to square the mean.

3CoreNon-calculator5 marks

The probability distribution of a discrete random variable $X$ is shown in the table, where $k$ is a constant.

$x$	$-2$	$0$	$3$	$5$
$P(X=x)$	$0.25$	$k$	$2k$	$0.15$

(a) Find the value of $k$. [2]

(b) Write down $P(X>0)$. [1]

(a) Uses sum of probabilities $=1$: $0.25+k+2k+0.15=1$M1
$k=0.2$A1
(b) $P(X>0)=0.55$ (FT their $k$: $2k+0.15$)A1
(c) Correct substitution into $\sum xP(X=x)$: $-2(0.25)+0(0.2)+3(0.4)+5(0.15)$ (FT their $k$)M1
$E(X)=1.45$ (accept $\frac{29}{20}$)A1
Answer: (a) $k=0.2$ (b) $0.55$ (c) $E(X)=1.45$
Examiner tip: Students often forget the negative value: $-2\times0.25$ must be subtracted, not added.

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4CoreCalculator7 marks

A gardener plants seeds from a packet of 20. Each seed germinates with probability $0.85$. Let $G$ be the number of seeds in the packet that germinate.

(a) State two assumptions needed for $G$ to be modelled by a binomial distribution. [2]

(b) Find the probability that exactly 18 seeds germinate. [2]

(d) Write down the expected number of seeds that germinate. [1]

(a) Seeds germinate independently of each other (accept 'whether one seed germinates does not affect another')R1
Probability of germinating is the same ($0.85$) for every seed. Do not accept 'fixed number of trials' alone as this is givenR1
(b) Recognises $G\sim B(20,\,0.85)$ and attempts $P(G=18)$, e.g. $\binom{20}{18}(0.85)^{18}(0.15)^2$(M1)
$0.229$ ($0.229338\ldots$)A1
(c) Attempts $P(G\le16)$ (e.g. binomcdf with upper bound 16); do not award for $P(G<16)$(M1)
$0.352$ ($0.352274\ldots$)A1
(d) $E(G)=17$A1
Answer: (b) $0.229$ (c) $0.352$ (d) $17$
Examiner tip: 'At most 16' means $P(G\le16)$; using $P(G\le15)$ or $1-P(G\le16)$ loses both marks.

5StandardNon-calculator7 marks

To play a game, a player pays $\$3$ and rolls a fair six-sided die once.

If the score is 6, the player receives $\$w$.
If the score is 4 or 5, the player receives $\$3$.
Otherwise the player receives nothing.

Let $G$ be the player's net gain in dollars (amount received minus the $\$3$ paid).

(a) Write down the probability distribution of $G$ as a table, giving the values of $G$ in terms of $w$ where necessary. [2]

(b) The game is fair. Find the value of $w$. [3]

(a) Values of $G$: $w-3,\ 0,\ -3$A1
with corresponding probabilities $\frac16,\ \frac13\ \left(=\frac26\right),\ \frac12\ \left(=\frac36\right)$A1
(b) Recognises fair game means $E(G)=0$(R1)
Correct equation, e.g. $\frac{w-3}{6}+0\cdot\frac13-3\cdot\frac12=0$ (FT their table). Accept equivalent method using amount received: $\frac{w}{6}+3\cdot\frac13=3$M1
$w=12$A1
(c) $E(G)=\frac{12}{6}-\frac32=0.5$ (FT from their (a))M1
Expected total gain $=60\times0.5=\$30$A1
Answer: (a) $G$: $w-3$, $0$, $-3$ with probabilities $\frac16,\frac13,\frac12$ (b) $w=12$ (c) $\$30$
Examiner tip: Fair means expected NET gain is zero; setting expected winnings equal to zero (forgetting the stake) gives the wrong $w$.

6StandardCalculator6 marks

A manufacturer finds that $7\%$ of its light bulbs are faulty. Bulbs are faulty independently of each other.

(a) A box contains 40 bulbs. Find the probability that the box contains at least one faulty bulb. [2]

(b) An inspector selects $n$ bulbs at random. Find the least value of $n$ such that the probability of selecting at least one faulty bulb is greater than $0.99$. [4]

(a) Uses $1-P(X=0)$ with $X\sim B(40,0.07)$, e.g. $1-0.93^{40}$(M1)
$0.945$ ($0.945132\ldots$)A1
(b) $P(\text{at least one})=1-0.93^n$(A1)
Sets up inequality $1-0.93^n>0.99$ (or $0.93^n<0.01$); accept equationM1
Valid attempt to solve, e.g. $n>\frac{\ln0.01}{\ln0.93}=63.457\ldots$, or table of values showing $n=63$ gives $0.98966$ and $n=64$ gives $0.99038$(A1)
$n=64$. Do not award for $n=63$ or $n=63.5$A1
Answer: (a) $0.945$ (b) $n=64$
Examiner tip: Dividing by $\ln0.93$, a negative number, reverses the inequality; also the answer must be the next whole number up, not the rounded value.

7StandardCalculator8 marks

The random variable $X$ follows a binomial distribution with $E(X)=4.8$ and $\mathrm{Var}(X)=3.84$.

(a) Find the parameters $n$ and $p$ of the distribution. [4]

(b) Find $P(X<E(X))$. [2]

(a) Writes $np=4.8$ and $np(1-p)=3.84$M1
Attempts to eliminate $n$, e.g. $1-p=\frac{3.84}{4.8}$M1
$p=0.2$A1
$n=24$A1
(b) Recognises $P(X\le4)$ for $X\sim B(24,0.2)$ (FT their $n$, $p$)(M1)
$0.460$ ($0.459877\ldots$)A1
(c) $\sigma=\sqrt{3.84}=1.9596\ldots$, so requires $P(X>6.7596\ldots)$, i.e. $P(X\ge7)$ or $1-P(X\le6)$(M1)
$0.189$ ($0.188928\ldots$)A1
Answer: (a) $n=24,\ p=0.2$ (b) $0.460$ (c) $0.189$
Examiner tip: $X$ takes only whole-number values, so $P(X<4.8)$ is $P(X\le4)$ and $P(X>6.76\ldots)$ is $P(X\ge7)$.

8HardCalculator6 marks

On a particular route, each of 8 buses on a weekday morning is late with probability $0.15$, independently of other buses.

A morning is classed as disrupted if at least 3 of the 8 buses are late.

(a) Find the probability that a randomly chosen morning is disrupted. [2]

(b) Find the probability that, in a 5-day working week, exactly 2 mornings are disrupted. [3]

(a) $L\sim B(8,0.15)$; uses $1-P(L\le2)$(M1)
$0.105$ ($0.105212\ldots$)A1
(b) Recognises second binomial $D\sim B(5,\,0.105212\ldots)$ (FT their (a))(M1)
Correct substitution, e.g. $\binom52(0.105212)^2(0.894787)^3$(A1)
$0.0793$ ($0.079304\ldots$). Accept $0.0792$–$0.0794$; using 3 s.f. value $0.105$ gives $0.0790$; award A0 for this premature roundingA1
(c) $5\times0.105212\ldots=0.526$ (FT)A1
Answer: (a) $0.105$ (b) $0.0793$ (c) $0.526$
Examiner tip: Keep the unrounded probability from (a) in the calculator for (b); rounding to 0.105 first changes the third significant figure.

9StretchCalculator7 marks

The bar chart shows the probability distribution of a random variable $X\sim B(10,\,p)$. The bars for $x=2$ and $x=3$ have exactly the same height.

(a) Show that $p=\dfrac{3}{11}$. [4]

(b) Write down the modes of $X$. [1]

(d) Find $P(X\ge5)$. [1]

(a) Uses $P(X=2)=P(X=3)$: $\binom{10}{2}p^2(1-p)^8=\binom{10}{3}p^3(1-p)^7$M1
Correct binomial coefficients $45$ and $120$ seen or usedA1
Divides by $p^2(1-p)^7$ (valid as $0<p<1$) to obtain $45(1-p)=120p$M1
$45=165p\Rightarrow p=\frac{45}{165}=\frac{3}{11}$ (AG). Must see a correct intermediate step before the given answerA1
(b) $2$ and $3$ (both required)A1
(c) $E(X)=\frac{30}{11}$ ($=2.73$)A1
(d) $P(X\ge5)=0.108$ ($0.107510\ldots$)A1
Answer: (b) 2 and 3 (c) $\frac{30}{11}\approx2.73$ (d) $0.108$
Examiner tip: In a 'show that', the working must reach $p=\frac{3}{11}$ without starting from it; substituting $\frac{3}{11}$ and checking equality earns at most the first two marks.

10StretchNon-calculatorHL only9 marks

A bag contains 4 red counters and 3 blue counters. Three counters are taken from the bag at random, without replacement. Let $X$ be the number of red counters taken.

(a) Explain why $X$ does not follow a binomial distribution. [1]

(b) Show that $P(X=2)=\dfrac{18}{35}$. [2]

A player pays $\$10$ to play and then receives $\$5$ for each red counter taken. Let $W$ be the player's net gain in dollars.

(d) Find $E(W)$ and $\mathrm{Var}(W)$. [2]

(a) The counters are not replaced, so the probability of red changes from one draw to the next (trials are not independent). Do not accept 'there are only 3 trials'R1
(b) Correct method, e.g. $3\times\frac47\times\frac36\times\frac35$ or $\frac{\binom42\binom31}{\binom73}$M1
$=\frac{108}{210}$ or $\frac{18}{35}$ (AG) with factor 3 (or combinations) clearly shownA1
(c) Full distribution: $P(X=0)=\frac1{35},\ P(X=1)=\frac{12}{35},\ P(X=3)=\frac4{35}$A1
$E(X)=\frac{12}{7}$ ($1.71$)A1
Attempts $E(X^2)-[E(X)]^2$ with $E(X^2)=\frac{120}{35}=\frac{24}{7}$M1
$\mathrm{Var}(X)=\frac{24}{49}$ ($0.490$)A1
(d) $W=5X-10$, $E(W)=5\cdot\frac{12}{7}-10=-\frac{10}{7}$ ($-\$1.43$) (FT their $E(X)$)A1
$\mathrm{Var}(W)=25\,\mathrm{Var}(X)=\frac{600}{49}$ ($12.2$). Award A0 if $-10$ affects the variance (FT their Var$(X)$)A1
Answer: (c) $E(X)=\frac{12}{7}$, $\mathrm{Var}(X)=\frac{24}{49}$ (d) $E(W)=-\frac{10}{7}$, $\mathrm{Var}(W)=\frac{600}{49}$
Examiner tip: Under a linear transformation the constant shifts the mean but not the variance, and the multiplier is squared: $\mathrm{Var}(5X-10)=25\,\mathrm{Var}(X)$.

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