A college has 80 students in Year 12 - Edexcel - A-Level Maths Statistics - Question 3 - 2015 - Paper 1

The probability can be calculated as:

1. Count the number of students studying only Chemistry:

   • Total studying Chemistry = 28
   • Students studying both Chemistry and Physics = 11
   • Students studying both Chemistry and Biology = 7
   • Students studying all three subjects = 3
   • Thus, students studying only Chemistry = 28 - (11 - 3) - (7 - 3) - 3 = 28 - 8 - 4 = 16

2. The required probability is then: $P(Chemistry \text{ but not Biology or Physics}) = \frac{16}{80} = 0.2$

We can find this probability using the formula for the union of two sets:

• Total students studying either Chemistry or Physics:
• Let ( P(C) ) be the number studying Chemistry, ( P(P) ) be the number studying Physics, and ( P(C \cap P) ) be the overlap (students studying both).

Using Inclusion-Exclusion: $P(C \cup P) = P(C) + P(P) - P(C \cap P)$ ( P(C) = 28 , P(P) = 30 , P(C \cap P) = 11 ) Thus, $P(C \cup P) = 28 + 30 - 11 = 47$

The probability: $P(C \cup P) = \frac{47}{80} = 0.5875$

From the previous calculations, we know:

• Total students = 80.
• Students studying Biology can be counted from those specified in the initial description:
  • Total studying Biology = 20.

Thus, the number of students not studying Biology is: $80 - 20 = 60$

The required probability is: $P(\text{not studying Biology}) = \frac{60}{80} = 0.75$

To determine if two events are independent, we check if: $P(A \cap B) = P(A) \cdot P(B)$ Where A is the event of studying Biology and B is the event of studying Chemistry.

Here,

• Number studying Biology = 20
• Number studying Chemistry = 28
• Number studying both = 7

Calculating,

• Probability of A: $P(A) = \frac{20}{80} = 0.25$
• Probability of B: $P(B) = \frac{28}{80} = 0.35$
• Probability of both A and B: $P(A \cap B) = \frac{7}{80} = 0.0875$

Now compare:

• $P(A) \cdot P(B) = 0.25 \cdot 0.35 = 0.0875$ Thus, since: $P(A \cap B) = P(A) \cdot P(B),$ Biology and Chemistry are statistically independent.