Probability: Venn; mutually exclusive — CJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
A group of 100 students are asked if they are student leaders, in a sports CCA, or studying in a science faculty. The number of students who are student leaders is 25, the number of students who are in a sports CCA is 30 and the number of students studying in a science faculty is 40. There are 15 student leaders who are in a sports CCA. The number of students who are student leaders, in a sports CCA and studying in a science faculty is \(x\). The number of students who are in a sports CCA and studying in a science faculty but not a student leader is \(y\).
One of the students is chosen at random.
\(A\) is the event that the student is a student leader.
\(B\) is the event that the student is in a sports CCA.
\(C\) is the event that the student is studying in a science faculty.
It is given that \(A\) and \(C\) are independent.
- Complete the Venn diagram below to represent all the above information. You are allowed to give expressions in terms of \(x\) and \(y\).
It is further given that \(B\) and \(C\) are independent.
- Find \(y\) in terms of \(x\). Hence, find the greatest and least possible values of \(y\).
Show full worked solution▾
(a) Since \(A\) and \(C\) are independent, \(\mathrm{P}(A\cap C) = \mathrm{P}(A)\times\mathrm{P}(C) = \dfrac{25}{100}\times\dfrac{40}{100} = \dfrac{10}{100}\), so \(\mathrm{n}(A\cap C)=10\).
Completed Venn diagram (numbers of students in each region):
(b) Since \(B\) and \(C\) are independent: \[\begin{aligned} \mathrm{P}(B\cap C) &= \mathrm{P}(B)\times\mathrm{P}(C) = \frac{30}{100}\times\frac{40}{100} = \frac{12}{100}\\ \mathrm{n}(B\cap C) &= 12\\ x+y &= 12\\ y &= 12-x \end{aligned}\]
Constraints: \(x\geq 0\), \(y\geq 0 \Rightarrow x\leq 12\). Also \(\mathrm{n}(A\cap B'\cap C) = 10-x\geq 0 \Rightarrow x\leq 10\).
Hence \(0\leq x\leq 10\), and \(y=12-x\).
Greatest value of \(y = 12\) (when \(x=0\)).
Least value of \(y = 2\) (when \(x=10\)).