What is the probability of rolling six ones in six rolls of a dice?
Karel's solution:
(1) The occurrences of a one in individual rolls are independent events.
(2) The probability of rolling a one of the first roll $(P_1)$ is $\frac16$, the probability of rolling a one on the second roll $(P_2)$ is $\frac16$, and so on, probability of rolling a one on the sixth roll $(P_6)$ is also $\frac16$.
(3) The probability of rolling six ones in six rolls $(P)$ is calculated as $P_1+P_2+\ldots+P_6=1$. This is a certain event!
Jan, Erica, Petr, Anna, and Barbara commented on his solution. Which of them, made a true statement about Karel's solution?
Jan: The mistake is in step (3). Rolling six ones in six rolls is the intersection of events "rolling one on the first roll," "rolling one on the second roll," and so on. The probability of the intersection of independent events is the product of their probabilities, so $P=\frac{1}{6^6}\cong0.0000214$.
Erica: The mistake is in step (3). The events are independent, so the sought probability does not depend on the number of rolls, i.e., $P=\frac16$.
Petr: The mistake is in step (1). The occurrences of a one in individual rolls are not independent events, since the probability of rolling a one increases with each subsequent roll. The correct way to calculate the probability of rolling six ones in six rolls is $P=\frac16\cdot\frac15\cdot\frac14\cdot\frac13\cdot\frac12\cdot1\cong0.00139$.
Barbara: The mistake is in step (3). Roling six ones in six rolls is the intersection of the events "rolling a one on the first roll," "rolling a one on the second roll," and so on. The probability of the intersection of independent events is the product of their probabilities. However, the result of the six rolls is not affected by the order in which the individual results occur. Therefore, we must multiply the probability by the number of permutations of the individual results. Thus, the probability is $P=6!\cdot\frac{1}{6^6}\cong0.0154$.
Anna: The solution is correct.