Jake and Kamil are sometimes late for school - Edexcel - A-Level Maths Statistics - Question 6 - 2011 - Paper 2

To find the probability that Kamil is late for school, we can use the formula relating probabilities:

$P(K) = P(K|J)P(J) + P(K|J')P(J')$

We need to find $P(K|J)$ and $P(K|J')$. From the given data:

• $P(J) = 0.25$
• $P(J ∩ K) = 0.15$
• $P(K) = P(J ∩ K) + P(J' ∩ K) = P(K) = 0.15 + P(J')P(K|J')$

Simplifying, we get:

1. $P(J') = 1 - P(J) = 0.75$
2. $P(K) = 0.15 + 0.75P(K|J')$
3. Solving for $P(K|J')$ using the initial condition, we find that:

$P(K) = 0.15 + 0.75(0.2) = 0.15 + 0.15 = 0.30$

Given that Jake is late for school, we need to find $P(K|J)$ using Bayes' theorem.
From the definitions, we derived that:

$P(K|J) = \frac{P(J ∩ K)}{P(J)} = \frac{0.15}{0.25} = 0.6$

To check for independence, we verify if:

$P(J ∩ K) = P(J)P(K)$

Using our previous findings:

• $P(J) = 0.25$

• $P(K)$ was derived as $0.30$. So we check:

• $P(J)P(K) = 0.25 imes 0.30 = 0.075$.

• But we know $P(J ∩ K) = 0.15$

Since $0.15$ does not equal $0.075$, J and K are not independent.

Since we determined that events J and K are not independent, this supports the teacher's suspicion that Jake being late for school and Kamil being late are linked in some way. The positive correlation suggests that external factors may influence both events simultaneously.