Two events A and B are such that P(A) = 0.2, P(A ∩ B) = 0.15 and P(A' ∩ B) = 0.6 - Leaving Cert Mathematics - Question 1 - 2010

To find the probability that neither A nor B happens, we denote this as $P(A' ∩ B')$:

$P(A' ∩ B') = 1 - (P(A) + P(B) - P(A ∩ B))$

Calculating this:

• $P(A) = 0.2$, $P(B) = 0.75$, and $P(A ∩ B) = 0.15$. Therefore:

$P(A' ∩ B') = 1 - (0.2 + 0.75 - 0.15) = 1 - 0.8 = 0.2$

Thus, the probability that neither A nor B happens is 0.2.

The conditional probability $P(A | B)$ is calculated using:

$P(A | B) = \frac{P(A ∩ B)}{P(B)}$

From previous calculations, we have:

• $P(A ∩ B) = 0.15$
• $P(B) = 0.75$

Thus:

$P(A | B) = \frac{0.15}{0.75} = 0.2$

Therefore, the conditional probability $P(A | B)$ is 0.2.

Two events A and B are independent if and only if:

$P(A ∩ B) = P(A) imes P(B)$

We already have:

• $P(A) = 0.2$,
• $P(B) = 0.75$,
• $P(A ∩ B) = 0.15$.

Calculating $P(A) imes P(B)$:

$P(A) imes P(B) = 0.2 imes 0.75 = 0.15$

Since $P(A ∩ B) = 0.15 = P(A) imes P(B)$, A and B are independent events.