Sure! Let's break down the problem step by step.
1. Intersection Calculation:
[tex]\[
|A \cap P| = 12 \cdot \binom{4}{2} \cdot 44
\][/tex]
In simpler terms, we have:
- 12 ways to choose a particular element.
- [tex]\(\binom{4}{2}\)[/tex] ways to choose 2 out of 4 (a common combinatorial calculation for choosing permutations).
- 44 ways to complete the scenario in our problem.
Multiplying these together gives us the size of the intersection between events [tex]\(A\)[/tex] and [tex]\(P\)[/tex]:
[tex]\[
|A \cap P| = 12 \cdot 6 \cdot 44 = 3168
\][/tex]
2. Total Ways to Choose Event [tex]\(A\)[/tex]:
[tex]\[
|A| = \binom{50}{3}
\][/tex]
This is the calculation of choosing 3 items out of a total of 50, which is a combinatorial calculation:
[tex]\[
|A| = 19600
\][/tex]
3. Probability Calculation:
The probability of [tex]\(P\)[/tex] given [tex]\(A\)[/tex] is represented as:
[tex]\[
p(P \mid A) = \frac{|A \cap P|}{|A|}
\][/tex]
Plugging in our calculated values:
[tex]\[
p(P \mid A) = \frac{3168}{19600}
\][/tex]
Simplifying this fraction (which we'll interpret approximately):
[tex]\[
p(P \mid A) \approx 0.1616
\][/tex]
Thus, the probability of [tex]\(P\)[/tex] given [tex]\(A\)[/tex] is approximately 0.1616.
