19. The probability of an event [tex]P[/tex] is [tex]\frac{3}{4}[/tex], while that of another event [tex]Q[/tex] is [tex]\frac{1}{6}[/tex]. If the probability of both [tex]P[/tex] and [tex]Q[/tex] is [tex]\frac{1}{12}[/tex], what is the probability of either [tex]P[/tex] or [tex]Q[/tex]?

A. [tex]\frac{1}{96}[/tex]
B. [tex]\frac{1}{8}[/tex]
C. [tex]\frac{5}{6}[/tex]
D. [tex]\frac{11}{12}[/tex]



Answer :

To find the probability of either event [tex]\(P\)[/tex] or event [tex]\(Q\)[/tex] occurring, we use the principle of inclusion-exclusion for probabilities. This principle states that the probability of either event [tex]\(P\)[/tex] or event [tex]\(Q\)[/tex] happening is given by:

[tex]\[ P(P \text{ or } Q) = P(P) + P(Q) - P(P \text{ and } Q) \][/tex]

We are given:
- The probability of event [tex]\(P\)[/tex], [tex]\(P(P) = \frac{3}{4}\)[/tex]
- The probability of event [tex]\(Q\)[/tex], [tex]\(P(Q) = \frac{1}{6}\)[/tex]
- The probability of both events [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] occurring at the same time, [tex]\(P(P \text{ and } Q) = \frac{1}{12}\)[/tex]

Now, substitute these values into the inclusion-exclusion formula:

[tex]\[ P(P \text{ or } Q) = \frac{3}{4} + \frac{1}{6} - \frac{1}{12} \][/tex]

To perform the addition and subtraction, we need a common denominator. The least common multiple of 4, 6, and 12 is 12. Therefore, we convert each fraction to have a denominator of 12:

- [tex]\(\frac{3}{4} = \frac{9}{12}\)[/tex] (since [tex]\(3 \times 3 = 9\)[/tex] and [tex]\(4 \times 3 = 12\)[/tex])
- [tex]\(\frac{1}{6} = \frac{2}{12}\)[/tex] (since [tex]\(1 \times 2 = 2\)[/tex] and [tex]\(6 \times 2 = 12\)[/tex])
- [tex]\(\frac{1}{12} = \frac{1}{12}\)[/tex]

Now we substitute these equivalents back into our formula:

[tex]\[ P(P \text{ or } Q) = \frac{9}{12} + \frac{2}{12} - \frac{1}{12} \][/tex]

Next, we combine the fractions:

[tex]\[ P(P \text{ or } Q) = \frac{9 + 2 - 1}{12} = \frac{10}{12} \][/tex]

Simplify the fraction [tex]\(\frac{10}{12}\)[/tex]:

[tex]\[ \frac{10}{12} = \frac{5}{6} \][/tex]

Therefore, the probability of either event [tex]\(P\)[/tex] or event [tex]\(Q\)[/tex] occurring is [tex]\(\frac{5}{6}\)[/tex].

The correct answer is:
C. [tex]\(\frac{5}{6}\)[/tex]