Selecting a group with combinatorics and conditional probability

Text

Class 4G has $21$ students; $4$ of them are drawn at random for an olympiad.

(a) In how many ways can the group be formed? Explain the formula.

(b) What is the probability that Lucia is selected?

(c) What is the probability that both Lucia and Marco are selected?

(d) What is the probability that at least one of Lucia and Marco is selected?

Solution

(a) The order of the draw does not matter and the students are all distinct, with no repetition: these are simple combinations of $21$ elements taken $4$ at a time: $\binom{21}{4} = 5985$

(b) The groups containing Lucia are obtained by fixing her and choosing the remaining $3$ among the other $20$ students: $P(\text{Lucia}) = \frac{\binom{20}{3}}{\binom{21}{4}} = \frac{1140}{5985} = \frac{4}{21} \approx 0.190$

(c) The groups containing both Lucia and Marco are obtained by fixing both and choosing the remaining $2$ among the other $19$ students: $P(\text{Lucia}\cap\text{Marco}) = \frac{\binom{19}{2}}{\binom{21}{4}} = \frac{171}{5985} = \frac{1}{35} \approx 0.0286$

(d) By the union formula (inclusion-exclusion principle), since $P(\text{Marco})=P(\text{Lucia})=\dfrac{4}{21}$: $P(\text{at least one}) = \frac{4}{21} + \frac{4}{21} - \frac{1}{35} = \frac{37}{105} \approx 0.352$

$\boxed{\binom{21}{4}=5985;\ P(L)=\tfrac{4}{21};\ P(L\cap M)=\tfrac{1}{35};\ P(\text{at least one})=\tfrac{37}{105}}$