IB Mathematics SL 4.6 Venn and tree diagrams, counting principles AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
(i) Find \( t \).
The number of students who play tennis is 19, including those who play only tennis (\( t \)) and those who play both tennis and volleyball (3).
\[ t + 3 = 19 \]
\[ t = 19 – 3 = 16 \]
(ii) Find \( v \).
Total students = 30. Students who play only tennis (\( t = 16 \)), both sports (3), neither (6), and only volleyball (\( v \)).
\[ t + 3 + v + 6 = 30 \]
\[ 16 + 3 + v + 6 = 30 \]
\[ v = 30 – 16 – 3 – 6 = 5 \]
Answer: (i) \( t = 16 \), (ii) \( v = 5 \).
Part (b)
Find the probability a student plays tennis or volleyball, but not both.
Students who play only tennis (\( t = 16 \)) or only volleyball (\( v = 5 \)).
Total = \( 16 + 5 = 21 \).
Probability: \[ \frac{21}{30} = \frac{7}{10} \]
Alternatively: Total playing tennis or volleyball = \( 19 + v – 3 = 19 + 5 – 3 = 21 \).
Exclude both: \( 21 – 3 = 18 \), but since we want exclusive, use only tennis or only volleyball: \( 16 + 5 = 21 \).
Answer: \( \frac{7}{10} \).
▶️ Answer/Explanation
Part (a)
Determine the values of \( p \), \( q \), and \( r \) in the Venn diagram.
(i) \( p \): Probability of \( A \cap B’ \).
\[ P(A) = P(A \cap B’) + P(A \cap B) \]
\[ 0.5 = p + 0.3 \]
\[ p = 0.5 – 0.3 = 0.2 \]
(ii) \( q \): Probability of \( B \cap A’ \).
\[ P(B) = P(B \cap A’) + P(A \cap B) \]
\[ 0.7 = q + 0.3 \]
\[ q = 0.7 – 0.3 = 0.4 \]
(iii) \( r \): Probability of \( A’ \cap B’ \).
\[ P(A \cup B) = P(A) + P(B) – P(A \cap B) \]
\[ P(A \cup B) = 0.5 + 0.7 – 0.3 = 0.9 \]
\[ P(A’ \cap B’) = 1 – P(A \cup B) = 1 – 0.9 = 0.1 \]
\[ r = 0.1 \]
Answer: (i) \( p = 0.2 \), (ii) \( q = 0.4 \), (iii) \( r = 0.1 \).
Part (b)
Find \( P(A|B’) \).
\[ P(A|B’) = \frac{P(A \cap B’)}{P(B’)} \]
From part (a): \[ P(A \cap B’) = p = 0.2 \]
\[ P(B’) = P(A \cap B’) + P(A’ \cap B’) = 0.2 + 0.1 = 0.3 \]
\[ P(A|B’) = \frac{0.2}{0.3} = \frac{2}{3} \]
Answer: \( P(A|B’) = \frac{2}{3} \).
Part (c)
Show that \( A \) and \( B \) are not independent.
For independence: \[ P(A \cap B) = P(A) \cdot P(B) \]
Given: \[ P(A \cap B) = 0.3 \]
\[ P(A) \cdot P(B) = 0.5 \cdot 0.7 = 0.35 \]
\[ 0.3 \neq 0.35 \]
Alternatively: \[ P(A|B’) = \frac{2}{3} \neq P(A) = 0.5 \]
Thus, \( A \) and \( B \) are not independent.
Answer: Since \( P(A \cap B) \neq P(A) \cdot P(B) \), events \( A \) and \( B \) are not independent.