Probability: Venn/tree, conditional, discrete & binomial, normal distribution, Bayes (HL), continuous random variables (HL)

1StandardCalculator7 marks

$A$ and $B$ are events. The Venn diagram shows the probabilities of the regions, where $x$ is the probability of being outside both events.

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

(b) Find $P(A\mid B)$. [2]

$0.35+0.15+0.25+x=1$M1
$x=0.25$A1
$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.15}{0.15+0.25}$M1
$0.375$ (accept $\tfrac38$)A1
Uses a valid independence test: compares $P(A\cap B)$ with $P(A)P(B)$ (or compares $P(A\mid B)$ with $P(A)$)M1
$P(A)=0.35+0.15=0.5$, $P(B)=0.15+0.25=0.4$, so $P(A)P(B)=0.2$ (or $P(A\mid B)=0.375$ and $P(A)=0.5$)A1
$P(A\cap B)=0.15\ne0.2$ (or $0.375\ne0.5$), so $A$ and $B$ are not independent (do not award R1 without both compared values seen)R1
Answer: (a) $x=0.25$ (b) $\tfrac38=0.375$ (c) not independent
Examiner tip: Test independence with $P(A\cap B)=P(A)P(B)$ using the full probabilities of $A$ and $B$ (including the overlap).

2StandardCalculator6 marks

A bag contains 5 red counters and 3 blue counters. Two counters are drawn at random, one after the other, without replacement.

(a) Find the probability that both counters are red. [2]

(b) Find the probability that the two counters are different colours. [2]

$\dfrac58\times\dfrac47$M1
$\dfrac{5}{14}$ $(=0.357)$A1
$P(RB)+P(BR)=\dfrac58\times\dfrac37+\dfrac38\times\dfrac57$M1
$\dfrac{15}{28}$A1
$P(R_1\mid\text{different})=\dfrac{P(RB)}{P(\text{different})}=\dfrac{15/56}{15/28}$M1
$\dfrac12$A1
Answer: (a) $\tfrac5{14}$ (b) $\tfrac{15}{28}$ (c) $\tfrac12$
Examiner tip: Without replacement the second denominator drops to 7; in (c) the denominator is the probability of the condition, not 1.

3StandardCalculator6 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.

4HardCalculator6 marks

The random variable $X\sim N(\mu,\sigma^2)$ is such that $P(X<20)=0.1$ and $P(X>35)=0.2$. Find the values of $\mu$ and $\sigma$.

Standardises: $P\left(Z<\dfrac{20-\mu}{\sigma}\right)=0.1$ and $P\left(Z<\dfrac{35-\mu}{\sigma}\right)=0.8$M1
$\dfrac{20-\mu}{\sigma}=-1.2816$A1
$\dfrac{35-\mu}{\sigma}=0.8416$A1
Subtracts/solves simultaneously: $15=(0.8416+1.2816)\sigma$M1
$\sigma=7.06$ ($7.0648\ldots$)A1
$\mu=29.1$ ($29.054\ldots$); accept $29.05$A1
Answer: $\sigma\approx7.06$, $\mu\approx29.1$
Examiner tip: Convert $P(X>35)=0.2$ to $P(X<35)=0.8$ before using inverse normal, and keep full accuracy until the end.

5StandardCalculator7 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.

6StandardCalculator6 marks

1 in 50 people in a population have a particular condition. A screening test gives a positive result for 95% of people who have the condition, and a (false) positive result for 8% of people who do not have it.

(a) Find the probability that a randomly chosen person tests positive. [3]

(b) A randomly chosen person tests positive. Find the probability that this person has the condition. [2]

Tree diagram or law of total probability: $P(+)=P(D)P(+\mid D)+P(D')P(+\mid D')$M1
$0.02\times0.95+0.98\times0.08$A1
$P(+)=0.0974$A1
$P(D\mid +)=\dfrac{P(D\cap +)}{P(+)}=\dfrac{0.02\times0.95}{0.0974}$ (follow through their $P(+)$ from (a))M1
$0.195$ ($0.19507\ldots$; accept exact $\tfrac{190}{974}=\tfrac{95}{487}$)A1
Valid comment linked to (b), e.g. only about 1 in 5 people who test positive actually have the condition / most positive results are false positives, because the condition is rare (accept: the probability is less than 0.5, so a positive result does not mean the person is likely to have the condition; the test alone is not reliable for diagnosis; a positive result should be followed up with further testing). Do not accept a comment that just repeats the number.R1
Answer: (a) $0.0974$ (b) $0.195$ (c) e.g. most positive results are false positives because the condition is rare
Examiner tip: Use the total probability of a positive (both branches) as the denominator in (b), not just the true-positive branch.

7HardCalculatorHL 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.

8HardCalculator8 marks

The mass of a bag of flour is normally distributed with mean $500$ g and standard deviation $8$ g. A bag is classed as underweight if its mass is less than $490$ g.

(a) Find the probability that a randomly chosen bag is underweight. [2]

(b) Five bags are chosen at random. Find the probability that exactly two are underweight. [3]

(c) The mean mass is adjusted to $\mu$ g, with the standard deviation unchanged, so that only $2\%$ of bags are underweight. Find $\mu$. [3]

$P(X<490)$ with $X\sim N(500,8^2)$ (e.g. $z=\dfrac{490-500}{8}=-1.25$ or GDC normal cdf)M1
$p=0.106$ ($0.105649\ldots$)A1
Recognises $Y\sim B(5,p)$ with $p$ from (a)M1
$\binom52p^2(1-p)^3$ (or GDC binomial pdf)M1
$0.0798$ ($0.079847\ldots$); a value of $0.0803$ from using $p=0.106$ loses this A1 onlyA1
$P(X<490)=0.02\Rightarrow\dfrac{490-\mu}{8}=-2.0537\ldots$M1
$\mu=490+8\times2.0537\ldots$A1
$\mu=506$ g ($506.43\ldots$; accept $506.4$)A1
Answer: (a) $0.106$ (b) $0.0798$ (c) $\mu=506$ g ($506.4$)
Examiner tip: Carry the unrounded value of $p$ from (a) into the binomial; a rounded $p$ can change the third significant figure.

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