Probability, Venn and tree diagrams, conditional probability and independence

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

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.

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4CoreNon-calculator6 marks

A fair four-sided spinner is numbered 1, 2, 3 and 4. A fair six-sided die is numbered 1 to 6. The spinner is spun and the die is rolled, and the score is the product of the two numbers obtained.

(a) Find the probability that the score is a multiple of 4. [2]

(b) The spinner is spun and the die is rolled 120 times. Find the expected number of times the score is a multiple of 4. [2]

Sample space of 24 equally likely outcomes (e.g. table/grid of products) with at least 8 correct favourable outcomes identifiedM1
$\dfrac{11}{24}$ (accept $0.458$)A1
$120\times\dfrac{11}{24}$ (follow through their (a))M1
$55$ (do not accept $55$ times rounded from an inexact probability unless it rounds correctly)A1
Identifies the 5 outcomes with score $>12$: $15, 16, 18, 20, 24$ (reduced sample space)M1
$\dfrac35$ (accept $0.6$)A1
Answer: (a) $\tfrac{11}{24}$ (b) $55$ (c) $\tfrac35$
Examiner tip: For a conditional probability from a sample space, count only within the restricted set: here the 5 outcomes above 12, not all 24.

5CoreCalculator10 marks

The table shows how 120 students in a sixth form travel to school.

	Walk	Bus	Car	Total
Year 12	15	32	13	60
Year 13	15	20	25	60
Total	30	52	38	120

A student is chosen at random from the 120 students. Let $T$ be the event that the student is in Year 13 and $W$ the event that the student walks.

(a) Find the probability that the student travels by car. [1]

(b) Find $P(T\cup W)$. [2]

(d) Show that $T$ and $W$ are independent. [3]

(e) Two different students are chosen at random from those who travel by bus. Find the probability that both are in Year 13. [2]

$\dfrac{38}{120}=\dfrac{19}{60}$ (accept $0.317$)A1
$\dfrac{60+15}{120}$ or $\dfrac{60}{120}+\dfrac{30}{120}-\dfrac{15}{120}$M1
$\dfrac{75}{120}=\dfrac58$ (accept $0.625$)A1
Denominator 52 seen, e.g. $\dfrac{32}{52}$M1
$\dfrac{8}{13}$ (accept $0.615$)A1
Valid test: compares $P(T\cap W)$ with $P(T)\times P(W)$ (or $P(W\mid T)$ with $P(W)$)M1
$P(T)=\dfrac{60}{120}=0.5$, $P(W)=\dfrac{30}{120}=0.25$ and $P(T\cap W)=\dfrac{15}{120}=0.125$ (or $P(W\mid T)=\dfrac{15}{60}=0.25$)A1
$0.5\times0.25=0.125=P(T\cap W)$ (or $P(W\mid T)=P(W)$), so $T$ and $W$ are independent AG (R1 only if both sides of the equality are shown numerically)R1
$\dfrac{20}{52}\times\dfrac{19}{51}$ (product of two fractions with denominators 52 and 51)M1
$\dfrac{95}{663}$ ($0.143$; $0.14328\ldots$)A1
Answer: (a) $\tfrac{19}{60}$ (b) $\tfrac58$ (c) $\tfrac{8}{13}$ (d) $0.5\times0.25=0.125=P(T\cap W)$ (e) $\tfrac{95}{663}\approx0.143$
Examiner tip: In (e) the sample space is the 52 bus users and the second student is drawn from 51 – using 120 or $(20/52)^2$ loses both marks.

6StandardNon-calculator7 marks

$A$ and $B$ are events such that $P(A)=0.4$, $P(B)=k$ and $P(A\cup B)=0.7$.

(a) Find $k$ if $A$ and $B$ are mutually exclusive. [2]

(b) Find $k$ if $A$ and $B$ are independent. [3]

Uses $P(A\cup B)=P(A)+P(B)$, i.e. $0.4+k=0.7$M1
$k=0.3$A1
Uses $P(A\cup B)=P(A)+P(B)-P(A\cap B)$M1
Substitutes $P(A\cap B)=0.4k$: $0.4+k-0.4k=0.7$M1
$k=0.5$A1
$P(A\cap B)=0.25k$ substituted: $0.4+k-0.25k=0.7$M1
$k=0.4$A1
Answer: (a) $0.3$ (b) $0.5$ (c) $0.4$
Examiner tip: Mutually exclusive means $P(A\cap B)=0$; independent means $P(A\cap B)=P(A)P(B)$ – do not mix the two up.

7StandardCalculator8 marks

80 students were asked which of Physics, Chemistry and Biology they study. The Venn diagram shows the number of students in each region.

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

A student is chosen at random from the 80 students.

(b) Find the probability that the student studies exactly one of the three subjects. [2]

(d) Two different students are chosen at random from the 80. Find the probability that both study exactly two of the subjects. [2]

$14+9+12+x+2x+6+5+4=80$ (or $50+3x=80$)M1
$x=10$A1
$14+9+12=35$ seen, over 80M1
$\dfrac{35}{80}=\dfrac{7}{16}$ (accept $0.4375$, $0.438$)A1
$n(\text{Biology})=12+6+20+5=43$ used as denominator (follow through their $x$)M1
$\dfrac{25}{43}$ ($0.581$)A1
Exactly two $=10+20+6=36$ and $\dfrac{36}{80}\times\dfrac{35}{79}$ (follow through their $x$; M0 for $\left(\tfrac{36}{80}\right)^2$)M1
$\dfrac{63}{316}$ ($0.199$; $0.19936\ldots$)A1
Answer: (a) $x=10$ (b) $\tfrac{7}{16}$ (c) $\tfrac{25}{43}\approx0.581$ (d) $\tfrac{63}{316}\approx0.199$
Examiner tip: "Exactly two" excludes the centre region (all three): students who include the 5 lose the mark.

8HardNon-calculator8 marks

A bag contains $n$ red counters and 3 blue counters, where $n>3$. Two counters are taken from the bag at random, without replacement.

(a) Write down, in terms of $n$, an expression for the probability that both counters are red. [1]

The probability that both counters are the same colour is $\dfrac12$.

(b) Show that $n^2-7n+6=0$. [4]

(d) Given that the two counters are the same colour, find the probability that they are both red. [2]

$\dfrac{n}{n+3}\times\dfrac{n-1}{n+2}$ (accept $\dfrac{n(n-1)}{(n+3)(n+2)}$)A1
Attempt at $P(\text{both blue})$ with denominators $n+3$ and $n+2$: $\dfrac{3}{n+3}\times\dfrac{2}{n+2}$M1
Adds both-red and both-blue and sets equal to $\tfrac12$: $\dfrac{n(n-1)+6}{(n+3)(n+2)}=\dfrac12$M1
Correctly cleared fractions: $2n^2-2n+12=n^2+5n+6$A1
Rearranges to $n^2-7n+6=0$ AG (no errors seen in the working)A1
$n=6$ only (the root $n=1$ must be rejected, as $n>3$; A0 if both values are given as the answer)A1
$\dfrac{P(\text{both red})}{P(\text{same})}=\dfrac{\frac{6}{9}\times\frac{5}{8}}{\frac12}$ (or $\dfrac{30}{30+6}$ from counting ordered pairs)M1
$\dfrac56$A1
Answer: (a) $\dfrac{n(n-1)}{(n+3)(n+2)}$ (c) $n=6$ (d) $\tfrac56$
Examiner tip: In a "show that", every step from the probability equation to the quadratic must appear – jumping straight to $n^2-7n+6=0$ scores no accuracy marks.

9HardCalculatorHL only8 marks

A company sends parcels using three couriers: $X$, $Y$ and $Z$. Courier $X$ handles 45% of the parcels, courier $Y$ handles 35% and courier $Z$ handles the rest. The percentages of parcels delivered late by $X$, $Y$ and $Z$ are 4%, 6% and 10% respectively.

A parcel is chosen at random. Let $L$ be the event that it is delivered late.

(a) Find $P(L)$. [2]

(b) Given that the parcel is delivered late, find the probability that it was handled by courier $Z$. [3]

$0.45\times0.04+0.35\times0.06+0.2\times0.1$ (three products summed; must use $P(Z)=0.2$)M1
$P(L)=0.059$A1
Uses Bayes' theorem / $\dfrac{P(Z\cap L)}{P(L)}$ with numerator $0.2\times0.1$M1
$\dfrac{0.02}{0.059}$ (follow through their (a))A1
$0.339$ ($0.33898\ldots$; accept $\tfrac{20}{59}$)A1
$P(L')=1-0.059=0.941$ (or $0.45\times0.96+0.35\times0.94+0.2\times0.9$)M1
$\dfrac{0.45\times0.96}{0.941}$ (numerator $0.432$)M1
$0.459$ ($0.45908\ldots$; accept $\tfrac{432}{941}$)A1
Answer: (a) $0.059$ (b) $\tfrac{20}{59}\approx0.339$ (c) $\tfrac{432}{941}\approx0.459$
Examiner tip: In (c) both the numerator and denominator must use the not-late branches ($0.96$ and $0.941$); reusing $0.04$ or $0.059$ is a common slip.

10StretchNon-calculator9 marks

Priya and Sam take turns to throw a ball at a target, with Priya throwing first. The first person to hit the target wins and the game stops. Each time she throws, Priya hits the target with probability $0.4$; each time he throws, Sam hits the target with probability $0.5$. All throws are independent.

(a) Find the probability that Sam wins with his first throw. [1]

(b) Find the probability that Priya wins with her second throw. [2]

(d) Given that Priya wins the game, find the probability that the game lasts no more than three throws. [2]

$0.6\times0.5=0.3$A1
$0.6\times0.5\times0.4$ (both miss first, then Priya hits)M1
$0.12$A1
Recognises that each round where both miss has probability $0.6\times0.5=0.3$M1
$P(\text{Priya wins})=0.4+0.3(0.4)+0.3^2(0.4)+\dots$ (infinite geometric series with at least three terms or general term $0.4(0.3)^{k-1}$)M1
Uses $S_\infty=\dfrac{u_1}{1-r}$ with $u_1=0.4$, $r=0.3$: $\dfrac{0.4}{1-0.3}$ (or equivalent method $p=0.4+0.3p$)M1
$=\dfrac{0.4}{0.7}=\dfrac47$ AGA1
$\dfrac{0.4+0.12}{4/7}$ (Priya can only win on throw 1 or throw 3; follow through their (b))M1
$0.91$ (accept $\tfrac{91}{100}$)A1
Answer: (a) $0.3$ (b) $0.12$ (c) $\dfrac{0.4}{1-0.3}=\dfrac47$ (d) $0.91$
Examiner tip: The repeating unit is a full round (both players miss, probability $0.3$), so the common ratio is $0.3$, not $0.6$.

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