Sure! Let's explore the concepts of conditional probability and the multiplication rule with regard to the probability of an event of the form "A and B." We can deduce some important properties of these probabilities from your question.
First, let's recap the context:
We are choosing a number randomly from the set of integers 1 through 10. We are interested in the probability of selecting a number that is both odd and prime. The goal is to calculate this probability using two different conditional probability approaches and confirm that both yield the same result.
### Step-by-Step Solution:
#### 1. Calculation Using \(P(\text{odd}) \cdot P(\text{prime} \mid \text{odd})\)
Step 1.1: Identify \(P(\text{odd})\):
- There are 5 odd numbers in the integers 1 through 10 (i.e., 1, 3, 5, 7, 9).
- Therefore,
[tex]\[
P(\text{odd}) = \frac{5}{10} = \frac{1}{2}.
\][/tex]
Step 1.2: Identify \(P(\text{prime} \mid \text{odd})\):
- Among the odd numbers 1, 3, 5, 7, and 9, the prime numbers are 3, 5, and 7 (which makes 3 primes out of 5 odd numbers).
- Therefore,
[tex]\[
P(\text{prime} \mid \text{odd}) = \frac{3}{5}.
\][/tex]
Step 1.3: Calculate the product [tex]\[ P(\text{odd}) \cdot P(\text{prime} \mid \text{odd})\][/tex]:
[tex]\[
P(\text{odd}) \cdot P(\text{prime} \mid \text{odd}) = \frac{5}{10} \cdot \frac{3}{5} = \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{10}.
\][/tex]
#### 2. Calculation Using \(P(\text{prime}) \cdot P(\text{odd} \mid \text{prime})\)
Step 2.1: Identify \(P(\text{prime})\):
- There are 4 prime numbers in the integers 1 through 10 (i.e., 2, 3, 5, 7).
- Therefore,
[tex]\[
P(\text{prime}) = \frac{4}{10} = \frac{2}{5}.
\][/tex]
Step 2.2: Identify \(P(\text{odd} \mid \text{prime})\):
- Among the prime numbers 2, 3, 5, and 7, the odd primes are 3, 5, and 7 (which makes 3 odd primes out of 4 prime numbers).
- Therefore,
[tex]\[
P(\text{odd} \mid \text{prime}) = \frac{3}{4}.
\][/tex]
Step 2.3: Calculate the product [tex]\[ P(\text{prime}) \cdot P(\text{odd} \mid \text{prime})\][/tex]:
[tex]\[
P(\text{prime}) \cdot P(\text{odd} \mid \text{prime}) = \frac{4}{10} \cdot \frac{3}{4} = \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}.
\][/tex]
### Conclusion:
Both methods produce the same result:
[tex]\[
\frac{3}{10}
\][/tex]
Thus, this consistency demonstrates an important property of conditional probability known as the multiplication rule. In general, for two events \(A\) and \(B\):
[tex]\[
P(A \text{ and } B) = P(A) \cdot P(B \mid A) = P(B) \cdot P(A \mid B).
\][/tex]
This rule arises from the properties of conditional probability and it signifies that the joint probability of two events [tex]\(A\)[/tex] and [tex]\(B\)[/tex] can be computed in either of the two equivalent ways. This reflects the underlying symmetry and connection between the events in terms of their probability.
