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]
[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]