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I've got a simple problem here, but I just want to ensure that I'm not losing a simple concept.

Relevant equations: $$P(A|B) = \frac{P(A \cap B)}{P(B)},$$

$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

Let $(S,P)$ be a sample space with $S = \{1, 2, 3, 4, 5\}$, and,
$P(1) = P(2) = 0.1$
$P(3) = P(4) = 0.2$
$P(5) = 0.4$

For each pair of events $A$ and $B$, find $P(A|B)$ and $P(B|A)$.

a. $A = \{1, 2, 3\}$; $B = \{2, 3, 4\}$

b. $A = \{1, 2, 3\}$; $B = \{4, 5\}$

c. $A = \emptyset$; $B = \{2, 3, 4\}$

d. $A = \{1, 2, 3, 4, 5\}; B = \{4, 5\}$

If I am asked to find the probability of one of these sets, do I just add the values together? Is there any instance I would multiply them; such as when I'm finding $P(A \cap B)$ vs. just $P(A)$ or $P(B)$? Do I need to use Inclusion-Exclusion?

Here's what I found.

a. $P(A|B) = \frac{0.3}{0.5} = \frac{3}{5}$, $P(B|A) = \frac{0.3}{0.4} = \frac{3}{4}$

b. $P(A|B) = 0$, $P(B|A) = 0$

c. $P(A|B) = 0$, $P(B|A) = undefined$

d. $P(A|B) = \frac{0.3}{0.3} = 1$, $P(B|A) = \frac{0.3}{1} = \frac{3}{5}$

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Your approach is correct, as are your answers. However, your solutions in part (d) are incorrect.

Since $A = \{1, 2, 3, 4, 5\}$ is the entire sample space, $P(A) = 1$. Since $P(4) = 0.2$ and $P(5) = 0.4$ and $B = \{4, 5\}$, $P(B) = P(4) + P(5) = 0.2 + 0.4 = 0.6$. Since $B \subseteq A$, $P(A \cap B) = P(B)$. Hence, \begin{align*} P(A \mid B) & = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = \frac{0.6}{0.6} = 1\\ P(B \mid A) & = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)} = \frac{0.6}{1} = 0.6 \end{align*}

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    $\begingroup$ Immaterial error corrected (took liberty). Corrected it from P(B) to P(A) in the second line P(B/A) $\endgroup$ – Satish Ramanathan Apr 26 '16 at 12:37

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