4.2 Probability

Q1

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M17/5/MATHL/HP1/ENG/TZ1/XX/4
Question Text	Six girls and five boys are seated randomly on a straight bench. Find the probability that the girls sit together and the boys sit together.
Total Mark	5
Correct Answer	1/231, 0.00433
Explanation	n/a
Mark Scheme	Step 1: Consider the given information. In this case, we can get the total number of arrangements that 11 students can sit on a bench. total number of arrangements = $11!$ We can also calculate the number of ways that girls and boys sit together. We multiply 2 because the boys group and girls group can change their order on the straight bench. number of ways for girls and boys to sit together = $6! \times 5! \times 2$ Step 2: Combine the given information. Hence, the probability can be calculated as below: $\frac{6! \times 5! \times 2}{11!} = \frac{1}{231}$

Q2

Topic	4.2 Probability
Tag	Probability Tree diagram Union De Morgan's Law Independent Events Mutually exclusive Addition law Bayes Theorem Conditional Experimental Theoretical Cumulative frequency
Source	N16-TZ0-P1-10(HL)
Question Text	Consider two events $A$ and $B$ defined in the same sample space. (a) Given that $P(A \cup B)=\frac{2}{3}, P(B \mid A)=\frac{1}{6}$ and $P\left(B \mid A^{\prime}\right)=\frac{1}{3}$ , (i) Find the value of $P(A)$ ; (a) $\frac{1}{4}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{2}{3}$ Total Mark: 5 Correct Answer: (c) Explanation: n/a Mark Scheme: Expanding $P(B \mid A)=\frac{1}{6}$ $P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{1}{6} \\ P(A \cap B)=\frac{P(A)}{6}$ Expanding $P\left(B \mid A^{\prime}\right)=\frac{1}{3}$ (ii) Find the value of $P(B)$ . (a) $\frac{1}{4}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{2}{3}$ Total Mark: 2 Correct Answer: (a) Explanation: n/a Mark Scheme: As $P(B)$ $= \frac{2-P(A)}6$ $P(B)$ $= \frac{2-(-0.5)}6=\frac14$

Q3

Topic	4.2 Probability
Tag
Source	M16-TZ1-P1-4(HL)
Question Text	Two events A and B are such that $P(A \cap B)=0.3$ and $P(A \cup B)=0.8$ . Find $P(A \mid B)$ (a) $\frac{3}{5}$ (b) $\frac{2}{5}$ (c) $\frac{3}{10}$ (d) $\frac{1}{10}$
Total Mark	4
Correct Answer	(b)
Explanation	n/a
Mark Scheme	Tip: Drawing a venn diagram may help for this question $P\left(A^{\prime} \mid B^{\prime}\right)=\frac{P(A \cap B)}{P(B)}=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}$ $P\left(B^{\prime}\right)=P\left(A \cap B^{\prime}\right)+1-P(A \cup B)$ $P(B')=0.3+0.2=0.5$ So, $P\left(A^{\prime} \mid B^{\prime}\right)=\frac{1-P(A \cup B)}{P(B')}=\frac{1-0.8}{0.5}=\frac{2}{5}$

Q4

Topic	4.2 Probability
Tag
Source	M16-TZ2-P1-7(HL)
Question Text	$A$ and $B$ are independent events such that $P(A)=1-P(B)=p, p \neq 0$ Find $P(A \mid A \cup B)$ in simplest form. (a) $\frac{1}{1+p^2}$ (b) $\frac{1}{2+p}$ (c) $\frac{p}{1+p^2}$ (d) $\frac{p}{1-p+p^2}$
Total Mark	6
Correct Answer	(d)
Explanation	n/a
Mark Scheme	$P(A \mid A \cup B)=\frac{P(A \cap(A \cup B))}{P(A \cup B)}=\frac{P(A)}{P(A \cup B)}$ Also, $P(A \cup B)=P(A)+P(B)-P(A \cap B)=P(A)+P(B)-P(A) P(B)$ So, $P(A \mid A \cup B)=\frac{P(A)}{P(A)+P(B)-P(A) P(B)}=\frac{p}{p+1-p-(p)(1-p)}$ $=\frac{p}{1-p+p^2}$

Q5

Topic	4.2 Probability
Tag
Source	N15-TZ0-P1-6(HL)
Question Text	A box contains three red balls and two blue balls. Adam and Bob play a game by each taking it in turn to take a ball from the box, without replacement. The first player to take a blue ball is the winner. (a) Adam plays first, find the probability that he wins. (a) $\frac25$ (b) $\frac35$ (c) $\frac{11}{20}$ (d) $\frac{13}{20}$
Total Mark	3
Correct Answer	(b)
Explanation	n/a
Mark Scheme	If the sequence of picks is represented with $R$ (red) and $B$ (blue), Adam wins when, $B, RRB$ So, $\frac25 + \left(\frac35 \times \frac24 \times \frac23\right) = \frac35$
Question Text	(b) The game is now changed so that the ball chosen is replaced after each turn. Find the probability of Adam winning (a) $\frac23$ (b) $\frac35$ (c) $\frac58$ (d) $\frac{7}{10}$
Total Mark	3
Correct Answer	(c)
Explanation	n/a
Mark Scheme	If the sequence of picks is represented with $R$ (red) and $B$ (blue), Adam wins when, B, RRB, RRRRB ... $\frac{2}{5}+\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)+\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)+\ldots=\frac{2}{5} \times \frac{1}{1-\frac{9}{25}}=\frac{5}{8}$ *note that this is an infinite geometric sequence

Q6

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M15-TZ1-P1-10(HL)
Question Text	A football team is playing a tournament of five matches. The probabilities that they win, draw or lose a match are $\frac23$ , $\frac{1}{12}$ and $\frac14$ respectively. These probabilities remain constant; the result of a match is independent of the results of other matches. At the end of the tournament, their coach John loses his job if they lose three consecutive matches, otherwise he does not lose his job. Find the probability that Roy loses his job. (a) $\frac{32}{81}$ (b) $\frac{37}{81}$ (c) $\frac{72}{283}$ (d) $\frac{88}{283}$
Total Mark	5
Correct Answer	(d)
Explanation	n/a
Mark Scheme	In order to lose three consecutive matches there must be either $3,4,5$ lost matches. Case 1: Lost three times $3 \times\left(\frac{2}{3}\right)^3 \times\left(\frac{1}{3}\right)^2=\frac{8}{81}$ $$ Case 2: Lost four times $2 \times\left(\frac{2}{3}\right)^4 \times\left(\frac{1}{3}\right)=\frac{32}{283}$ Case 3: Lost five times $\left(\frac{2}{3}\right)^4=\frac{32}{283}$ \end{gathered} So in total $\frac{8}{81}+\frac{32}{283}+\frac{32}{283}=\frac{88}{283}$

Q7

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M15-TZ2-P1-1(HL)
Question Text	$A$ and $B$ are two events such that $P(A) = 0.5, P(B) = 0.4,$ and $P(A∪B) = 0.7$ . (a) Find $100 \times P(A∩B)$
Total Mark	2
Correct Answer	20
Explanation	n/a
Mark Scheme	$P(A∪B) = P(A) + P(B) - P(A∩B)\\ P(A∩B) = 0.5 + 0.4 - 0.7 = 0.2\\ 100 \times P(A∩B) = 20$
Question Text	(b) Determine whether events $A$ and $B$ are independent. (a) Dependent (b) Independent
Total Mark	2
Correct Answer	(b)
Explanation	n/a
Mark Scheme	n/a

Q8

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	N14/5/MATHL/HP1/ENG/TZ0/XX/4
Question Text	Events $A$ and $B$ are such that $P(A) = 0.3$ and $P(B) = 0.4$ (a) Determine the value of $100 \times P(A∪B)$ when (i) $A$ and $B$ are mutually exclusive; Total Mark: 2 Correct Answer: 70 Explanation: n/a Mark Scheme: use of $P(A∪B)=P(A)+P(B)$ $P(A∪B)=0.3+0.4=0.7$ (ii) $A$ and $B$ are independent. Total Mark: 3 Correct Answer: 58 Explanation: n/a Mark Scheme: $P(A∪B)=P(A)+P(B)-P(A)P(B)$ $P(A∪B)=0.3+0.4-0.12=0.58$
Question Text	(b) Determine the range of possible values of $P(A\|B)$ . (a) $0 ≤ P(A\|B) ≤ 0.75$ (b) $0.4 ≤ P(A\|B) ≤ 1$ (c) $0.3 ≤ P(A\|B) ≤ 0.4$ (d) $0.3 ≤ P(A\|B) ≤ 1$
Total Mark	4
Correct Answer	(a)
Explanation	n/a
Mark Scheme	$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$ $P(A \mid B)$ is a maximum when $P(A \cap B)=P(A)$ $\frac{P(A \cap B)}{P(B)}=\frac{0.3}{0.4}=0.75$ $P(A \mid B)$ is a minimum when $P(A \cap B)=0$ $0 \leq P(A \mid B) \leq 0.75$

Q9

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M14/5/MATHL/HP1/ENG/TZ2/XX/1
Question Text	Events $A$ and $B$ are such that $P(A) = \frac 25, P(B) = \frac3{10}$ and $P(A\|B)= \frac {10}{21}$ (a) Find $a + b, P(A∩B) = \frac ab$ where $a,b$ are positive integers in lowest terms
Total Mark	3
Correct Answer	8
Explanation	n/a
Mark Scheme	$P(A∩B) = P(A\|B) \times P(B)$ $P(A∩B) = \frac3{10} \times \frac{10}{21}$ $= \frac17$
Question Text	(b) Find $c + d, P(A∩B) = \frac cd$ where $c,d$ are positive integers in lowest terms
Total Mark	2
Correct Answer	109
Explanation	n/a
Mark Scheme	$P(A∪B) = P(A) + P(B) - P(A∩B)$ $= \frac25 + \frac3{10} - \frac17 = \frac{39}{70}$
Question Text	(c) State whether $A$ and $B$ are independent (a) Dependent (b) Independent
Total Mark	2
Correct Answer	(a)
Explanation	n/a
Mark Scheme	$P(A∩B) ≠ P(A) \times P(B)$ So not independent

Q10

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M14/5/MATHL/HP1/ENG/TZ2/XX/11
Question Text	Mobile phone batteries are produced by two machines. Machine A produces 70% of the daily output and machine B produces 30%. It is found by testing that on average 3% of batteries produced by machine A are faulty and 1% of batteries produced by machine B are faulty (a) A battery is selected at random. Find the probability that it is faulty. (a) 0.015 (b) 0.018 (c) 0.021 (d) 0.024
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	$P(Faulty) = 0.7 × 0.03 + 0.3 × 0.01$ $= 0.024$
Question Text	(b) A battery is selected at random and found to be faulty. Find the probability that was produced by machine A. (a) $\frac{7}{8}$ (b) $\frac{15}{16}$ (c) $\frac34$ (d) $\frac{4}{5}$
Total Mark	4
Correct Answer	(a)
Explanation	n/a
Mark Scheme	Probability of A given that it is faulty. $P(A\|Faulty) = \frac{P(A∩F)}{P(F)}$ $=\frac{0.7×0.03}{0.024} = \frac{0.021}{0.024} = \frac78$

Q11

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	N13-TZ0-P1-2(HL)
Question Text	The discrete random variable $X$ has probability distribution:

$x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac14$	$\frac13$	$\frac1{12}$	$a$

Question Text	(a) Find the value of $a$ (A) $\frac16$ (B) $\frac13$ (C) $\frac13$ (D) $\frac14$
Total Mark	1
Correct Answer	(c)
Explanation	n/a
Mark Scheme	$\frac14 + \frac13 + \frac1{12} + a = 1$ $a = \frac13$
Question Text	(b) Find $E(X)$ . (A) $\frac32$ (B) $\frac43$ (C) $\frac{17}{12}$ (D) $2$
Total Mark	2
Correct Answer	(a)
Explanation	n/a
Mark Scheme	$E(X) = \frac13 + \left(2 × \frac1{12}\right) + \left(3 × \frac13\right) = \frac32$
Question Text	(c) Find $Var(X)$ . (A) $1$ (B) $\frac43$ (C) $\frac{17}{12}$ (D) $2$
Total Mark	3
Correct Answer	(c)
Explanation	n/a
Mark Scheme	$\begin{gathered}E\left(X^2\right)=\frac{1}{3}+2^2 \times \frac{1}{12}+3^2 \times \frac{1}{3}=\frac{11}{3} \\ \text { Using } \operatorname{Var}(X)=E\left(X^2\right)-(E(X))^2 \\ =\frac{11}{3}-\frac{9}{4}=\frac{17}{12}\end{gathered}$

Q12

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M13-TZ1-P1-9(HL)
Question Text	Two events $A$ and $B$ are such that $P(A∪B) = 0.6$ and $P(A'∣B') = 0.5.$ Find $100 \times P(B)$ .
Total Mark	6
Correct Answer	20
Explanation	n/a
Mark Scheme	$P(A'∣B') = \frac{P(A'∩B') }{ P(B')} = \frac{1 - P(A∪B)}{P(B')}$ $P(A'∩B') = P(A∪B)' = 1 - 0.6 = 0.4$ $0.5 = \frac{0.4}{ P(B')}$ $P(B') = 0.8$ $P(B) = 0.2$

Q13

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M13-TZ2-P1-4(HL)
Question Text	Brian and Chris buy a box of 20 chocolates of which 12 are milk and 8 are dark. Brian randomly takes a chocolate and eats it. Then Chris randomly takes a chocolate and eats it. The probability that Brain and Chris eat the same type of chocolate can be expressed as $\frac ab$ where $a, b$ are positive integers in the lowest terms. Find the value of $a + b$ .
Total Mark	4
Correct Answer	142
Explanation	n/a
Mark Scheme	Case 1: Both eat white $\frac{12}{20} \times \frac{11}{19} = \frac{33}{95}$ Case 2: Both eat dark $\frac{8}{20} \times \frac{7}{19} = \frac{14}{95}$ Total probability is therefore, $\frac{47}{95}$

Q14

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	N18/5/MATHL/HP1/ENG/TZ0/XX/1
Question Text	Consider two events, $A$ and $B$ , such that $P(A) = P(A'∩B) = 0.3$ . Find $100 \times P(A∪B)$ .
Total Mark	3
Correct Answer	60
Explanation	n/a
Mark Scheme	$P(A∪B) = P(A) + P(A'∩B) = 0.6$

Q15

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M18/5/MATHL/HP1/ENG/TZ1/XX/3
Question Text	Two unbiased tetrahedral (four-sided) dice with faces labelled 2, 3, 4, 5 are thrown and the scores recorded. Let the random variable $t$ be the maximum of these two scores. The probability distribution of $T$ is given in the following table.

$t$	$0$	$1$	$2$	$3$
$P(T=t)$	$\frac1{16}$	$\frac3{16}$	$a$	$\frac7{16}$

Question Text	(a) Find the value of $a$ (A) $\frac3{16}$ (B) $\frac14$ (C) $\frac{5}{16}$ (D) $\frac7{16}$
Total Mark	2
Correct Answer	(c)
Explanation	n/a
Mark Scheme	$a=1-\frac1{16}-\frac3{16}-\frac7{16}=\frac5{16}$
Question Text	(b) Find the expected value of $T$ . (A) $3$ (B) $4$ (C) $\frac{29}{8}$ (D) $\frac{33}{8}$
Total Mark	2
Correct Answer	(d)
Explanation	n/a
Mark Scheme	$E(T) = \frac{2+9+20+35}{16} = \frac{33}8$

Q16

Topic	4.2 Probability
Tag	Probability; Tree diagram; Union; De Morgan's Law; Independent; Events; Mutually exclusive; Addition law; Bayes Theorem; Conditional; Experimental; Theoretical; Cumulative frequency
Source	M18/5/MATHL/HP1/ENG/TZ2/XX/3
Question Text	The discrete random variable $X$ has the following probability distribution, where $p$ is a constant.

$x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$p$	$0.2$	$0.3$	$0.41-p$	$p^2$

Question Text	(a) Find the value of $p$ (A) 0.25 (B) 0.3 (C) 0.35 (D) 0.4
Total Mark	2
Correct Answer	(b)
Explanation	n/a
Mark Scheme	Equating sum of probabilities to 1 $p+0.2+0.3+0.41-p+p^2=1 \\ p^2=0.09, p=0.3$
Question Text	(b) (i) Find $\mu$ , the expected value of $X$ (A) 1.59 (B) 1.6 (C) 1.61 (D) 1.62 Total Mark: 2 Correct Answer: (a) Explanation: n/a Mark Scheme: $\begin{aligned} & \mu=0 \times 0.3+1 \times 0.2+2 \times 0.3+3 \times 0.11+4 \times 0.09 \\ & =1.59\end{aligned}$ (ii) Find $P(X>\mu)$ (A) 0.2 (B) 0.3 (C) 0.4 (D) 0.5 Total Mark: 4 Correct Answer: (d) Explanation: n/a Mark Scheme: $\begin{aligned} & P(X>\mu)=P(X=2)+P(X=3)+P(X=4) \\ & \quad=0.5\end{aligned}$