To solve for [tex]\( P(A \text{ or } B) \)[/tex], which refers to the probability that a randomly chosen student likes either pepperoni or olives (or both), we use the formula from probability theory:
[tex]\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \][/tex]
Given:
- [tex]\( P(A) = \frac{2}{3} \)[/tex], which is the probability that a student likes pepperoni.
- [tex]\( P(B) = \frac{1}{6} \)[/tex], which is the probability that a student likes olives.
We also need to consider the probability that a student likes both pepperoni and olives, [tex]\( P(A \text{ and } B) \)[/tex]. Since it is not provided, we assume this probability to be negligible (or zero) in our calculations:
[tex]\[ P(A \text{ or } B) = P(A) + P(B) \][/tex]
Now, substituting the values:
[tex]\[ P(A \text{ or } B) = \frac{2}{3} + \frac{1}{6} \][/tex]
To add these fractions, we need a common denominator. The common denominator is 6:
[tex]\[ \frac{2}{3} = \frac{4}{6} \][/tex]
So,
[tex]\[ P(A \text{ or } B) = \frac{4}{6} + \frac{1}{6} = \frac{5}{6} \][/tex]
Converting [tex]\(\frac{5}{6}\)[/tex] to decimal gives approximately [tex]\(0.8333\)[/tex], which matches our calculated value.
Thus, the answer is not directly one of the options given in your list. If we consider [tex]\( \frac{5}{6} \)[/tex] in its numerical form (or approximately [tex]\(0.83\)[/tex]), the closest option would be the one closest to this value.
However, based on our given options, it seems there's a mistake or rounding issue, so none of the options completely match [tex]\(\frac{5}{6}\)[/tex]. If we were to correct this issue, we should select the closest fraction from standard options, which here are incorrectly suited. Therefore, with correct interpretation, the answer should ideally be [tex]\(\frac{5}{6}\)[/tex].
