Bayes' theorem and continuous random variables – IB Maths exam questions with mark scheme

1HardCalculatorHL only8 marks

The continuous random variable $X$ has probability density function $f(x)=k(4-x)$ for $0\le x\le4$, and $f(x)=0$ otherwise.

(a) Show that $k=\dfrac18$. [2]

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

(d) Find $P(X>3)$. [1]

$\int_0^4k(4-x)\,dx=1$: $k\left[4x-\tfrac{x^2}2\right]_0^4=1$M1
$k\left(16-8\right)=8k=1$, so $k=\dfrac18$ AGA1
$E(X)=\int_0^4\dfrac{x(4-x)}8\,dx$ (limits required)M1
$\dfrac43$A1
Sets $\int_0^m\dfrac{4-x}8\,dx=\dfrac12$ (or $\int_m^4\dfrac{4-x}8\,dx=\dfrac12$)M1
$\dfrac{4m-\tfrac12m^2}{8}=\dfrac12$, i.e. $m^2-8m+8=0$A1
$m=4-2\sqrt2\ (\approx1.17)$, rejecting $4+2\sqrt2$ as it lies outside $0\le x\le4$A1
$P(X>3)=\int_3^4\dfrac{4-x}8\,dx=\dfrac1{16}$ (accept $0.0625$)A1
Answer: (a) $8k=1$ shown (b) $\tfrac43$ (c) $4-2\sqrt2\approx1.17$ (d) $\tfrac1{16}$
Examiner tip: Reject the root of the median equation that lies outside the domain; and write the integral limits explicitly.

2CoreCalculatorHL only6 marks

A phone repair shop buys replacement screens from three suppliers, $A$, $B$ and $C$. Of all screens bought, $50\%$ come from $A$, $30\%$ from $B$ and $20\%$ from $C$. The probability that a screen is faulty is $0.02$ for supplier $A$, $0.03$ for supplier $B$ and $0.05$ for supplier $C$.

A screen is chosen at random.

(a) Find the probability that the screen is faulty. [2]

(b) Given that the screen is faulty, find the probability that it came from supplier $C$. [2]

Attempt to sum three products: $0.5(0.02)+0.3(0.03)+0.2(0.05)$M1
$P(F)=0.029$A1
Recognising conditional probability with their (a) as denominator: $\dfrac{0.2\times0.05}{0.029}$M1
$\dfrac{10}{29}=0.345$ (3 s.f.)A1
$\dfrac{0.5\times0.98}{1-0.029}$, i.e. $\dfrac{0.49}{0.971}$ (follow through their (a))M1
$0.505$ (3 s.f.) (accept $\tfrac{490}{971}$)A1
Answer: (a) $0.029$ (b) $\tfrac{10}{29}\approx0.345$ (c) $\approx0.505$
Examiner tip: In (c) the numerator must use $P(\text{not faulty}\mid A)=0.98$, not $0.02$, and the denominator is $1-0.029$.

3CoreNon-calculatorHL only7 marks

The discrete random variable $X$ has the probability distribution shown, where $a$ and $b$ are constants.

$x$	0	1	2	3
$P(X=x)$	$0.1$	$a$	$b$	$0.2$

It is given that $E(X)=1.6$.

(a) Find the value of $a$ and the value of $b$. [3]

(b) Find $\mathrm{Var}(X)$. [2]

Uses $\sum P=1$: $a+b=0.7$M1
Uses $\sum xP(X=x)=1.6$: $a+2b+0.6=1.6$M1
$a=0.4,\ b=0.3$A1
$E(X^2)=1(0.4)+4(0.3)+9(0.2)=3.4$ (follow through their $a,b$)M1
$\mathrm{Var}(X)=3.4-1.6^2=0.84$A1
$E(5-2X)=5-2(1.6)=1.8$A1
$\mathrm{Var}(5-2X)=(-2)^2(0.84)=3.36$ (follow through their (b); A0 for $-2\times0.84$ or $-1.68$)A1
Answer: (a) $a=0.4,\ b=0.3$ (b) $0.84$ (c) $1.8$, $3.36$
Examiner tip: Adding a constant does not change the variance, and the multiplier is squared: $\mathrm{Var}(5-2X)=4\,\mathrm{Var}(X)$, never negative.

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4StandardNon-calculatorHL only7 marks

The continuous random variable $X$ has probability density function $f$ shown in the diagram. The graph of $f$ is the horizontal line $y=h$ for $0\le x\le2$ and a straight line from $(2,h)$ to $(4,0)$ for $2\le x\le4$. $f(x)=0$ otherwise.

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

(b) Find $P(1<X<3)$. [3]

Total area $=1$: $2h+\tfrac12(2)(h)=1$M1
$h=\dfrac13$A1
Area from 1 to 2: $1\times\tfrac13=\tfrac13$A1
Area from 2 to 3 found, e.g. trapezium $\tfrac12\left(\tfrac13+\tfrac16\right)(1)=\tfrac14$, or $\displaystyle\int_2^3\frac{4-x}{6}\,dx$M1
$P(1<X<3)=\dfrac7{12}$A1
Recognises median lies in $[0,2]$ (area to $x=2$ is $\tfrac23>\tfrac12$) and sets $\tfrac13m=\tfrac12$M1
$m=1.5$A1
Answer: (a) $\tfrac13$ (b) $\tfrac7{12}$ (c) $1.5$
Examiner tip: Check which section contains the median before setting up an equation: here the first rectangle already holds two-thirds of the probability.

5StandardCalculatorHL only10 marks

The continuous random variable $X$ has probability density function

$$f(x)=\begin{cases}kx^2(3-x), & 0\le x\le3\\ 0, & \text{otherwise.}\end{cases}$$

(a) Show that $k=\dfrac4{27}$. [2]

(b) Find the mode of $X$. [2]

(d) The median of $X$ is $1.84$, correct to 3 significant figures. With reference to your answers, comment on the shape of the distribution. [2]

$\displaystyle k\int_0^3(3x^2-x^3)\,dx=1$ with correct integration $k\left[x^3-\tfrac{x^4}4\right]_0^3$M1
$k\cdot\tfrac{27}4=1\Rightarrow k=\tfrac4{27}$ (AG; must see $\tfrac{27}4$)A1
Sets $f'(x)=0$: $k(6x-3x^2)=0$ (or uses GDC maximum of graph)M1
Mode $=2$A1
$\displaystyle E(X)=\int_0^3 x f(x)\,dx=1.8$A1
Attempt $\displaystyle\int_0^3x^2f(x)\,dx$ $(=3.6)$M1
$\mathrm{Var}(X)=3.6-1.8^2$ $(=0.36)$M1
$\sigma=0.6$A1
States mean $<$ median $<$ mode (i.e. $1.8<1.84<2$)R1
Hence negatively skewed (tail to the left)A1
Answer: (a) shown (b) $2$ (c) $E(X)=1.8$, $\sigma=0.6$ (d) mean $<$ median $<$ mode, negative skew
Examiner tip: Standard deviation is the square root of the variance: candidates who stop at $0.36$ lose the final mark.

6StandardNon-calculatorHL only7 marks

Sam travels to college by bus, by bicycle or on foot. On any given day, the probability that Sam takes the bus is $\tfrac12$, cycles is $\tfrac13$ and walks is $\tfrac16$. The probability that Sam is late is $\tfrac15$ if travelling by bus, $\tfrac1{10}$ if cycling and $\tfrac14$ if walking.

(a) Show that the probability that Sam is late on a randomly chosen day is $\tfrac7{40}$. [2]

(b) On one day Sam is late. Find the probability that Sam walked. [2]

(d) Hence determine whether the events “Sam takes the bus” and “Sam is late” are independent. Justify your answer. [2]

$\tfrac12\cdot\tfrac15+\tfrac13\cdot\tfrac1{10}+\tfrac16\cdot\tfrac14$M1
$\tfrac{12}{120}+\tfrac4{120}+\tfrac5{120}=\tfrac{21}{120}=\tfrac7{40}$ (AG; common denominator or equivalent must be seen)A1
$\dfrac{\tfrac16\times\tfrac14}{\tfrac7{40}}$M1
$\dfrac5{21}$A1
$\dfrac{\tfrac1{10}}{\tfrac7{40}}=\dfrac47$A1
Compares $P(B\mid L)=\tfrac47$ with $P(B)=\tfrac12$ (or $P(B\cap L)=\tfrac1{10}$ with $P(B)P(L)=\tfrac7{80}$)R1
Not independent, as $\tfrac47\ne\tfrac12$ (do not award without a valid comparison)A1
Answer: (a) shown (b) $\tfrac5{21}$ (c) $\tfrac47$ (d) not independent
Examiner tip: To test independence you must compare two numbers explicitly, e.g. $P(B\mid L)$ with $P(B)$; a statement alone scores nothing.

7HardNon-calculatorHL only10 marks

The continuous random variable $X$ has probability density function $f(x)=\cos x$ for $0\le x\le\dfrac{\pi}2$, and $f(x)=0$ otherwise.

(a) Show that $f$ is a valid probability density function. [2]

(b) Find the exact value of the median of $X$. [2]

(d) Given that $\displaystyle\int_0^{\pi/2}x^2\cos x\,dx=\frac{\pi^2}4-2$, show that $\mathrm{Var}(X)=\pi-3$. [3]

$\cos x\ge0$ for $0\le x\le\tfrac{\pi}2$R1
$\displaystyle\int_0^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=1$A1
$\displaystyle\int_0^m\cos x\,dx=\tfrac12\Rightarrow\sin m=\tfrac12$M1
$m=\dfrac{\pi}6$ (reject other solutions outside the domain)A1
Attempt by parts with $u=x$, $dv=\cos x\,dx$: $[x\sin x]-\int\sin x\,dx$M1
$[x\sin x+\cos x]_0^{\pi/2}$A1
$\left(\tfrac{\pi}2+0\right)-(0+1)=\tfrac{\pi}2-1$ (AG)A1
Uses $\mathrm{Var}(X)=E(X^2)-[E(X)]^2$ with $E(X^2)=\tfrac{\pi^2}4-2$M1
$\left(\tfrac{\pi}2-1\right)^2=\tfrac{\pi^2}4-\pi+1$A1
$\tfrac{\pi^2}4-2-\tfrac{\pi^2}4+\pi-1=\pi-3$ (AG)A1
Answer: (a) shown (b) $\tfrac{\pi}6$ (c) shown (d) shown
Examiner tip: A valid pdf needs two conditions: $f(x)\ge0$ throughout and total area 1; showing only the integral loses the first mark.

8HardNon-calculatorHL only9 marks

The continuous random variable $X$ has probability density function $f(x)=ax+b$ for $0\le x\le2$, and $f(x)=0$ otherwise, where $a$ and $b$ are constants. It is given that $E(X)=\dfrac76$.

(a) Find the value of $a$ and the value of $b$. [4]

(b) Find $\mathrm{Var}(X)$. [3]

$\displaystyle\int_0^2(ax+b)\,dx=1\Rightarrow 2a+2b=1$M1
$\displaystyle\int_0^2(ax^2+bx)\,dx=\tfrac76\Rightarrow\tfrac{8a}3+2b=\tfrac76$M1
Attempt to solve simultaneouslyM1
$a=\tfrac14,\ b=\tfrac14$A1
Attempt $\displaystyle E(X^2)=\int_0^2\frac{x^3+x^2}4\,dx$M1
$E(X^2)=\tfrac53$A1
$\mathrm{Var}(X)=\tfrac53-\tfrac{49}{36}=\tfrac{11}{36}$A1
$E(Y)=6\cdot\tfrac76-2=5$A1
$\mathrm{Var}(Y)=36\cdot\tfrac{11}{36}=11$ (follow through their (b))A1
Answer: (a) $a=b=\tfrac14$ (b) $\tfrac{11}{36}$ (c) $E(Y)=5$, $\mathrm{Var}(Y)=11$
Examiner tip: The condition on $E(X)$ gives the second equation; the first always comes from total probability $=1$.

9StretchCalculatorHL only8 marks

A screening test is used for a rare condition. A proportion $p$ of a population has the condition. If a person has the condition, the test is positive with probability $0.95$. If a person does not have the condition, the test is positive with probability $0.1$.

(a) Show that the probability that a randomly chosen person tests positive is $0.1+0.85p$. [2]

(b) Health officials find that half of the people who test positive actually have the condition. Find the value of $p$. [3]

A person who tests positive is given the test a second time. Assume that results of the two tests are independent for a given person.

$0.95p+0.1(1-p)$M1
$=0.95p+0.1-0.1p=0.1+0.85p$ (AG)A1
Uses Bayes: $\dfrac{0.95p}{0.1+0.85p}=0.5$M1
$0.95p=0.05+0.425p$ (or any correct rearranged linear equation)A1
$p=\dfrac2{21}=0.0952$ (3 s.f.)A1
Numerator $0.95^2p$ (or $0.5\times0.95$ using updated prior)M1
Denominator $0.95^2p+0.1^2(1-p)$ (or $0.5\times0.95+0.5\times0.1$)M1
$\dfrac{19}{21}=0.905$ (3 s.f.)A1
Answer: (a) shown (b) $p=\tfrac2{21}\approx0.0952$ (c) $\tfrac{19}{21}\approx0.905$
Examiner tip: Both positive results must be multiplied in each branch: $0.95^2$ for the condition and $0.1^2$ for no condition, not $0.95\times0.1$.

10StretchCalculatorHL only10 marks

The continuous random variable $X$ has probability density function

$$f(x)=\begin{cases}kx^2, & 0\le x\le1\\ \dfrac{k(3-x)}2, & 1<x\le3\\ 0, & \text{otherwise,}\end{cases}$$

where $k$ is a positive constant. The graph of $y=f(x)$ is shown.

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

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

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

(e) Find $P(X<2\mid X>1)$. [2]

$\displaystyle\int_0^1kx^2\,dx+\int_1^3\frac{k(3-x)}2\,dx=1$, i.e. $\tfrac k3+k=1$M1
$k=\tfrac34$A1
Mode $=1$A1
$P(X\le1)=\tfrac14<\tfrac12$ so median in $1<x\le3$R1
$\displaystyle\int_m^3\frac{3(3-x)}8\,dx=\tfrac12$ (or $\tfrac14+\int_1^m\ldots=\tfrac12$), i.e. $\tfrac3{16}(3-m)^2=\tfrac12$M1
$m=3-\sqrt{\tfrac83}=1.37$ (3 s.f.) (reject $4.63$)A1
$\displaystyle\int_0^1\tfrac34x^3\,dx+\int_1^3\tfrac38x(3-x)\,dx$M1
$E(X)=\tfrac{23}{16}=1.4375$ (accept $1.44$)A1
$\dfrac{P(1<X<2)}{P(X>1)}$ with $P(X>1)=\tfrac34$M1
$\dfrac{9/16}{3/4}=\dfrac34$A1
Answer: (a) $\tfrac34$ (b) $1$ (c) $1.37$ (d) $\tfrac{23}{16}$ (e) $\tfrac34$
Examiner tip: For a piecewise pdf, find the cumulative probability at the join first so you integrate the correct piece for the median.

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