Probability - Properties of conditional probability

If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find (i) $P(A \cap B)$ (ii) $P(A|B)$ (iii) $P(A \cup B)$.

(i) Using the multiplication rule: $P(A \cap B) = P(A) \cdot P(B|A) = 0.8 \times 0.4 = 0.32$ (ii) Using the definition of conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = 0.64$ (iii) Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.8 + 0.5 - 0.32 = 0.98$

We first find the intersection using the given conditional probability, then use that intersection to find the reverse conditional probability and the union probability.

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Let $F$ be the event that the sum is 6: $F = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$. So, $n(F) = 5$. Let $E$ be the event that 4 appears at least once: $E = \{(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)\}$. The intersection $E \cap F = \{(2,4), (4,2)\}$. Therefore, $n(E \cap F) = 2$. The required probability is: $P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{n(E \cap F)}{n(F)} = \frac{2}{5}$

By restricting the sample space to event $F$ (sum is 6), we calculate the probability of $E$ occurring within that subset.

If $P(A|B) = 0.3$ and $P(B) = 0.5$, find $P(A'|B)$.

Using Property 3 of conditional probability: $P(A'|B) = 1 - P(A|B)$ $P(A'|B) = 1 - 0.3 = 0.7$

The probability of the complement of an event under the same condition is simply 1 minus the probability of the event.