Question

now, find the probability of event c.

\\(p(c) = \frac{?}{5}\\)

Explanation:

Identify the paths leading to event C

To find the total probability of event \(C\), we identify all paths in the tree diagram that end in \(C\).

• Path 1: First branch to \(A\), then branch to \(C\).
• Path 2: First branch to \(B\), then branch to \(C\).

Calculate the probability of each path

We multiply the probabilities along each branch for both paths:

• Probability of Path 1:

$$ P(A \cap C) = P(A) \cdot P(C|A) = \frac{1}{4} \cdot \frac{3}{5} = \frac{3}{20} $$

• Probability of Path 2:

$$ P(B \cap C) = P(B) \cdot P(C|B) = \frac{3}{4} \cdot \frac{3}{5} = \frac{9}{20} $$

Sum the path probabilities

We add the probabilities of the two mutually exclusive paths to find \(P(C)\):

$$ P(C) = P(A \cap C) + P(B \cap C) = \frac{3}{20} + \frac{9}{20} = \frac{12}{20} $$

Simplify the fraction to match the target format

The target format is \(P(C) = \frac{[?]}{5}\). We simplify the fraction \(\frac{12}{20}\) by dividing the numerator and the denominator by 4:

$$ P(C) = \frac{12 \div 4}{20 \div 4} = \frac{3}{5} $$

Thus, the missing value in the numerator is 3.

Answer:

Now, find the probability of event C.
\(P(C) =\) <blank>\(\frac{3}{5}\)</blank>