IB Mathematics SL 4.6 Use of Venn diagrams, tree diagrams AI HL Paper 1- Exam Style Questions- New Syllabus
Question
In a city, 32% of people have blue eyes. If someone has blue eyes, the probability that they also have fair hair is 58%. This information is represented in the following tree diagram.
(a) Write down the value of \( a \). [1]
(b) Find an expression, in terms of \( b \), for the probability of a person not having blue eyes and having fair hair. [1]
It is known that 41% of people in this city have fair hair.
(c)(i) Calculate the value of \( b \). [2]
(c)(ii) Calculate the value of \( c \). [1]
▶️ Answer/Explanation
Markscheme
(a)
From the tree diagram, \( a = 1 – 0.58 = 0.42 \). A1
Value of \( a \): 0.42.
[1 mark]
From the tree diagram, \( a = 1 – 0.58 = 0.42 \). A1
Value of \( a \): 0.42.
[1 mark]
(b)
Probability of not blue eyes and fair hair: \( P(B’ \cap F) = 0.68 \times b \). A1
Expression: \( 0.68 b \).
[1 mark]
Probability of not blue eyes and fair hair: \( P(B’ \cap F) = 0.68 \times b \). A1
Expression: \( 0.68 b \).
[1 mark]
(c)(i)
Total probability of fair hair:
\[ \begin{aligned} P(\text{fair hair}) &= 0.32 \times 0.58 + 0.68 \times b \\ &= 0.41 \end{aligned} \] M1
Solve:
\[ \begin{aligned} 0.1856 + 0.68 b &= 0.41 \\ 0.68 b &= 0.41 – 0.1856 \\ b &\approx 0.33 \end{aligned} \] A1
Value of \( b \): 0.33.
[2 marks]
Total probability of fair hair:
\[ \begin{aligned} P(\text{fair hair}) &= 0.32 \times 0.58 + 0.68 \times b \\ &= 0.41 \end{aligned} \] M1
Solve:
\[ \begin{aligned} 0.1856 + 0.68 b &= 0.41 \\ 0.68 b &= 0.41 – 0.1856 \\ b &\approx 0.33 \end{aligned} \] A1
Value of \( b \): 0.33.
[2 marks]
(c)(ii)
From the tree diagram, \( c = 1 – b = 1 – 0.33 = 0.67 \). A1
Value of \( c \): 0.67.
[1 mark]
From the tree diagram, \( c = 1 – b = 1 – 0.33 = 0.67 \). A1
Value of \( c \): 0.67.
[1 mark]
Total Marks: 5
Question
Lucas analyzes data from 130 applicants who applied for admission into either the Arts programme or the Sciences programme at a university. The outcomes of their applications are shown in the following table.
| Programme | Accepted | Rejected |
|---|---|---|
| Arts | 17 | 24 |
| Sciences | 25 | 64 |
(a) Find the probability that a randomly chosen applicant from this group was accepted by the university. [1]
An applicant is chosen at random from this group. It is found that they were accepted into the programme of their choice.
(b) Find the probability that the applicant applied for the Arts programme. [2]
Two different applicants are chosen at random from the original group.
(c) Find the probability that both applicants applied to the Arts programme. [3]
▶️ Answer/Explanation
Markscheme
(a)
Total applicants: 130. Total accepted: \( 17 + 25 = 42 \).
\[ \begin{aligned} P(\text{accepted}) &= \frac{17 + 25}{130} \\ &= \frac{42}{130} = \frac{21}{65} \approx 0.323076 \end{aligned} \] A1
[1 mark]
Total applicants: 130. Total accepted: \( 17 + 25 = 42 \).
\[ \begin{aligned} P(\text{accepted}) &= \frac{17 + 25}{130} \\ &= \frac{42}{130} = \frac{21}{65} \approx 0.323076 \end{aligned} \] A1
[1 mark]
(b)
Given the applicant was accepted, total accepted: 42. Arts accepted: 17.
\[ \begin{aligned} P(\text{Arts} | \text{accepted}) &= \frac{17}{17 + 25} \\ &= \frac{17}{42} \approx 0.404761 \end{aligned} \] A1 A1
[2 marks]
Given the applicant was accepted, total accepted: 42. Arts accepted: 17.
\[ \begin{aligned} P(\text{Arts} | \text{accepted}) &= \frac{17}{17 + 25} \\ &= \frac{17}{42} \approx 0.404761 \end{aligned} \] A1 A1
[2 marks]
(c)
Total Arts applicants: \( 17 + 24 = 41 \).
Probability both apply to Arts:
\[ \begin{aligned} P(\text{both Arts}) &= \frac{41}{130} \times \frac{40}{129} \\ &= \frac{41 \times 40}{130 \times 129} = \frac{1640}{16770} \\ &\approx 0.0977936 \end{aligned} \] A1 M1 A1
[3 marks]
Total Arts applicants: \( 17 + 24 = 41 \).
Probability both apply to Arts:
\[ \begin{aligned} P(\text{both Arts}) &= \frac{41}{130} \times \frac{40}{129} \\ &= \frac{41 \times 40}{130 \times 129} = \frac{1640}{16770} \\ &\approx 0.0977936 \end{aligned} \] A1 M1 A1
[3 marks]
Total Marks: 6