To find the intersection probability [tex]\(P(A \cap B)\)[/tex] given [tex]\(P(A) = \frac{1}{3}\)[/tex], [tex]\(P(B) = \frac{2}{5}\)[/tex], and [tex]\(P(A \cup B) = \frac{3}{5}\)[/tex], we can use the formula for the union of two events.
The formula for the union of two events [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is given by:
[tex]\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\][/tex]
We need to find [tex]\(P(A \cap B)\)[/tex], so we will rearrange the formula to solve for [tex]\(P(A \cap B)\)[/tex]:
[tex]\[P(A \cap B) = P(A) + P(B) - P(A \cup B)\][/tex]
Substitute the given probabilities into the formula:
[tex]\[P(A \cap B) = \frac{1}{3} + \frac{2}{5} - \frac{3}{5}\][/tex]
First, find the common denominator to add [tex]\(\frac{1}{3}\)[/tex] and [tex]\(\frac{2}{5}\)[/tex]:
The least common multiple of 3 and 5 is 15.
Convert the fractions to have a common denominator:
[tex]\[\frac{1}{3} = \frac{5}{15}\][/tex]
[tex]\[\frac{2}{5} = \frac{6}{15}\][/tex]
Add these fractions together:
[tex]\[\frac{5}{15} + \frac{6}{15} = \frac{5 + 6}{15} = \frac{11}{15}\][/tex]
Now, subtract [tex]\(\frac{3}{5}\)[/tex] from [tex]\(\frac{11}{15}\)[/tex], keeping in mind to convert [tex]\(\frac{3}{5}\)[/tex] to have the common denominator:
[tex]\[\frac{3}{5} = \frac{9}{15}\][/tex]
So we need to perform the following subtraction:
[tex]\[\frac{11}{15} - \frac{9}{15} = \frac{11 - 9}{15} = \frac{2}{15}\][/tex]
Therefore, the probability [tex]\(P(A \cap B)\)[/tex] is [tex]\(\frac{2}{15}\)[/tex].
The answer is [tex]\(\boxed{\frac{2}{15}}\)[/tex].
