In a class of 36 students, each student is either in a sports team or a performing arts club. Each student is also either a member or is a leader in their respective co-curricular activities. One student is selected at random from the class. The probability that the student selected is in a sports team is $\dfrac{1}{3}$.

1. Find the total number of students in sports teams.

Number of students in sports teams
$= \frac{1}{3} \times 36 = 12$

2. Find the number of leaders in sports teams.

Given that there are $4$ sports team members,
$\text{number of sports team leaders} = 12 - 4 = 8$

1. Let the number of students who are members in a performing arts club be an unknown.

Let the number of students who are members in a performing arts club be $x$.

2. Form the probability expression for selecting two such students without replacement.

Since two students are selected at random from $36$ students without replacement,
$\frac{x}{36}\times \frac{x-1}{35}=\frac{1}{42}.$

3. Solve the resulting quadratic equation.

$\frac{x(x-1)}{36\times 35}=\frac{1}{42}$
$x(x-1)=\frac{36\times 35}{42}=30$
$x^2-x-30=0$
$(x-6)(x+5)=0$
So $x=6$ or $x=-5$.

4. Reject the impossible negative value and state the valid number of students.

The number of students cannot be negative, so $x=6$.

1. Explain why adding the two probabilities directly is not generally correct.

Let $L$ be the event that a student is a leader and $P$ be the event that a student is in a performing arts club.

For “leader or performing arts club”,
$P(L\cup P)=P(L)+P(P)-P(L\cap P).$
So
$P(L\cup P)=k+m-P(L\cap P).$
Since a student can be both a leader and in a performing arts club, Derrick's claim $k+m$ is not necessarily correct.