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 need to directly use the provided probability statement:

$P(K) = P(J ∩ K) + P(J' ∩ K)$ Knowing that:

• $P(J ∩ K) = 0.15$
• To find: $P(K) = P(J ext{ late}) + P(J' ext{ late})$ You can rearrange the use of the joint probability to find $P(K)$ if needed (but it's directly not provided, so we take a value).

However, with given context: Conversely, if we assume area is covered with known probabilities, we have: $P(K) = P(J ext{ late}) - 0.15$ Considering we can derive equivalently via totals, thus: The probability that Kamil is late for school is to be calculated by knowing other events' statistics.

Given that Jake is late for school, we want to calculate: $P(K | J) = \frac{P(J ∩ K)}{P(J)}$ Substituting in the values:

• $P(J ∩ K) = 0.15$
• $P(J) = 0.25$

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

Therefore, the probability that Kamil is late given that Jake is late is 0.6.

To determine independence, we check: $P(J ∩ K) = P(J) imes P(K)$ From our previous findings, we note:

• $P(J) = 0.25$
• And assuming we derive K as say later from supports in b) and c) yielding relations together for checks if totaling outer values yield true or needing corrections. Since we have interaction: The independence is noted by these relations equating to 0.075 (multiplying independent factors) which does not equate to joint found. Hence, Jake and Kamil's lateness are not independent.

The calculations indicate that the lateness of Jake and Kamil are not statistically independent. This confirms the teacher's suspicion that Jake being late is linked to Kamil being late, as their probabilities do not act separately—rather, they are influenced by each other. Therefore, the teacher's suspicion appears valid given the calculations.