Using the Principle of Inclusion-Exclusion, calculate the number of students who played all three games (football, netball, and basketball) given the following data. Total students = 100, |F|=40, |N|=

The number of students who played at least one game is the total number of students minus those who played none.

We use the Principle of Inclusion-Exclusion:

Substitute the known values:

So, 8 students played all three games.

• Students playing exactly F and N: $|F \cap N| - |F \cap N \cap B| = 24 - 8 = 16$
• Students playing exactly F and B: $|F \cap B| - |F \cap N \cap B| = 18 - 8 = 10$
• Students playing exactly B and N: $|B \cap N| - |F \cap N \cap B| = 19 - 8 = 11$ Total students playing exactly two games $= 16 + 10 + 11 = 37$.

We can find this by subtracting the number of students playing exactly two games and exactly three games from the total number of students playing at least one game.

So, 37 students played exactly one game.

a) Basing on the calculation from the information given, advise whether the school should offer sports bursaries.

Condition 1: More than 60% play at least one of the games.
Number of students playing at least one game $= 82$.
Percentage $= \frac{82}{100} \times 100\% = 82\%$.
Since $82\% > 60\%$, this condition is met.

Condition 2: More than 40% must know how to play at least two of the three games.
Number of students playing at least two games = (Exactly two games) + (Exactly three games)
Number of students playing at least two games $= 37 + 8 = 45$.
Percentage $= \frac{45}{100} \times 100\% = 45\%$.
Since $45\% > 40\%$, this condition is met.

Since both conditions are met, the school should offer sports bursaries.

b) Calculate the probability that a student picked at random played only one sport game.

The number of students who played only one sport game is 37.
The total number of students is 100.