Find the missing probability.

[tex]\[ P(A) = \frac{11}{20}, \quad P(B \mid A) = \frac{13}{20}, \quad P(A \cap B) = \text{?} \][/tex]

A. [tex]\(\frac{143}{400}\)[/tex]

B. [tex]\(\frac{4}{25}\)[/tex]

C. [tex]\(\frac{121}{200}\)[/tex]

D. [tex]\(\frac{61}{100}\)[/tex]



Answer :

To find the missing probability [tex]\(P(A \cap B)\)[/tex], we will use the concept of conditional probability. The formula for the conditional probability of event [tex]\(B\)[/tex] given event [tex]\(A\)[/tex] is:

[tex]\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \][/tex]

Rearranging the formula to solve for [tex]\(P(A \cap B)\)[/tex], we have:

[tex]\[ P(A \cap B) = P(A) \cdot P(B \mid A) \][/tex]

Given:
- [tex]\( P(A) = \frac{11}{20} \)[/tex]
- [tex]\( P(B \mid A) = \frac{13}{20} \)[/tex]

Now, we multiply these probabilities:

[tex]\[ P(A \cap B) = \frac{11}{20} \cdot \frac{13}{20} \][/tex]

We calculate the product:

[tex]\[ P(A \cap B) = \frac{11 \cdot 13}{20 \cdot 20} \][/tex]
[tex]\[ P(A \cap B) = \frac{143}{400} \][/tex]

We now compare this result with the given options:

A. [tex]\(\frac{143}{400}\)[/tex]

B. [tex]\(\frac{4}{25} = \frac{16}{100} = 0.16\)[/tex]

C. [tex]\(\frac{121}{200} = 0.605\)[/tex]

D. [tex]\(\frac{61}{100} = 0.61\)[/tex]

The correct probability [tex]\( P(A \cap B) \)[/tex] is therefore option A: [tex]\(\frac{143}{400}\)[/tex].