Conditional Probability Problems about Die Rolling
Problem 728
A fair six-sided die is rolled.
(1) What is the conditional probability that the die lands on a prime number given the die lands on an odd number?
(2) What is the conditional probability that the die lands on 1 given the die lands on a prime number?
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Solution.
Let $E$ and $F$ be events. Then the conditional probability that $E$ occurs given $F$ has occurred, which is denoted by $P(E\mid F)$, is defined as
\[P(E \mid F) = \frac{P(E \cap F)}{P(F)}.\]
For uniform space (such as fair die rolling), this can be rewritten as
\[P(E \mid F) = \frac{|E \cap F|}{|F|}.\]
Here the notation $|A|$ means the number of elements in $A$.
Solution (1)
Let $E = \{2, 3, 5\}$ be the set of prime number faces of a die. Let $F = \{1, 3, 5\}$ be the set of odd faces of a die. Then the required conditional probability can be calculated as follows:
\begin{align*}
P(E \mid F) &= \frac{|E \cap F|}{|F|}\\
&= \frac{|\{2, 3, 5\} \cap \{1, 3, 5\}|}{|\{1, 3, 5\}|}\\
&= \frac{|\{3, 5\}|}{\{1,3,5\}|}\\
&= \frac{2}{3}
\end{align*}
Solution (2)
The required conditional probability is
\begin{align*}
P(\{1\} \mid E) &= \frac{|\{1\} \cap E|}{|E|}\\
&= \frac{|\{1\} \cap \{2, 3, 5\}|}{|\{2, 3, 5\}|}\\
&= \frac{|\emptyset|}{|\{2, 3, 5\}|}\\
&= \frac{0}{3} = 0
\end{align*}
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