A large college produces three magazines - Edexcel - A-Level Maths Statistics - Question 4 - 2021 - Paper 1

From the Venn diagram and noting that no students read all three magazines, we can express P(G ∩ E' ∩ S') (which is p) using the probabilities given in the diagram:

Using the equation:

$P(G) = 0.25 = 0.08 + 0.05 + q + p$

The value of p can be determined as follows: Since P(G) = 0.25 and we already have values for other parts:

$p = 0.25 - (0.08 + 0.05 + q)$

Substituting 0 for q:

$p = 0.25 - 0.13 = 0.12$

As established earlier, we know:

$P(G ∩ E ∩ S) = 0$ where q is the overlapping probability between G and E. To find q, we use the equation derived from the total probabilities:

Using the earlier established values:

$P(G) = 0.25 = 0.08 + 0.05 + q + p$

Since we already determined that p = 0.12:

Substituting in the probability of p gives us:

$q = 0.25 - (0.08 + 0.05 + 0.12) = 0.25 - 0.25 = 0.12$

To find r, we utilized the conditional probability provided:

We know that:

P(S | E) = rac{P(S ∩ E)}{P(E)} Given that P(S | E) = 5/12: Substituting to solve for P(S):

We know the total probabilities:

Using previously established values we can set: $P(E) = 0.12 + q = 0.12 + 0.12 = 0.24$ Then, substituting back yields:

ightarrow r = 0.10 $$

To find t, we need to understand that the sum of all probabilities must equal 1:

$1 = 0.08 + 0.05 + 0.09 + q + p + r + t$ Where we can substitute in the values:

$1 = 0.08 + 0.05 + 0.09 + 0.12 + 0.10 + t$ Solving for t will give:

$t = 1 - (0.08 + 0.05 + 0.09 + 0.12 + 0.10) = 0.20$

To determine independence, we utilize the independence condition:

Events A and B are independent if and only if:

$P(A ∩ B) = P(A) imes P(B)$ Here: Let A = (S ∩ E') and B = G. First, we need to compute:

$P(S ∩ E') = 0.36 \text{ (not part of E)}$ Given the independence condition, we find:

ightarrow 0.36 ot\equiv 0.25 $$ Thus, S ∩ E' and G are not independent.