The Venn diagram shows the probabilities associated with four events, A, B, C and D - Edexcel - A-Level Maths Statistics - Question 1 - 2020 - Paper 1

Using the formula for the total probability of events in the Venn diagram:

$P(B) = P(A ext{ extbf{only}}) + P(B ext{ extbf{only}}) + 0.24 + p + 0.16$

We have:

0.4 = 0.07 + 0.24 + p + 0.16 \\ 0.4 = 0.47 + p \\ \Rightarrow p = 0.4 - 0.47 = -0.07 $$ Since a probability cannot be negative, the calculation needs correction: Subtracting the fixed probabilities: $$ p = 0.4 - (0.24 + 0.07 + 0.16) = 0.09 $$

The independence of events A and B implies:

$P(A \cap B) = P(A) P(B)$

From the Venn diagram, since we don't know P(A) directly, we write:

$P(A) (0.4) = 0.24 \implies P(A) = \frac{0.24}{0.4} = 0.6$

Since we know:

$P(A) = q + 0.16 + 0.24 = 0.6$

This implies:

q + 0.40 = 0.6 \\ \Rightarrow q = 0.6 - 0.40 = 0.20 $$

To find the value of r, we use the conditional probability formula:

$P(B'|C) = \frac{P(B' \cap C)}{P(C)} \rightarrow 0.64 = \frac{r}{r+p}$

This leads us to:

0.64 (0.09 + r) = r \\ 0.064 + 0.64r = r \\ 0.064 = r - 0.64r \\ 0.064 = 0.36r \\ \Rightarrow r = \frac{0.064}{0.36} \approx 0.16 $$

To find the value of s, we can use the total probability rule:

From the Venn diagram, the expression for total probability says:

$P(A) + P(B) + P(C) + P(D) = q + 0.16 + 0.09 + s = 1$

We already know:

1. $P(A) = 0.6$,
2. $P(C)= 0.09 + 0.16 + r = 0.25$,
3. $P(B) = 0.4$,
4. etc. Thus, we have:

0.49 + s = 1 \\ \Rightarrow s = 1 - 0.49 = 0.51 $$