Question

A random experiment consists of throwing 100 fair dice, each die having six faces numbered 1 to 6. An event $A$ represents the set of all outcomes where at least one of the dice shows a 1. Then, $P(A) =$

• A. 0
• B. 1
• C. $1 - \left(\frac{5}{6}\right)^{100}$
• D. $\left(\frac{5}{6}\right)^{100}$

Hint:

When calculating the probability of "at least one" in independent events, it is often easier to first calculate the probability of the complement event (i.e., "none of the events occur") and then subtract it from 1.

Answer 1

The Correct Option is C

We are asked to find the probability that at least one of the 100 dice shows a 1 when thrown.

The probability of getting a 1 on a single die is $\frac{1}{6}$, and the probability of not getting a 1 (i.e., getting one of the other five faces) on a single die is $\frac{5}{6}$.

Step 1: Probability of no die showing a 1
We first calculate the probability that none of the 100 dice shows a 1. Since the dice rolls are independent, the probability that a single die does not show a 1 is $\frac{5}{6}$. Therefore, the probability that none of the 100 dice shows a 1 is:
$$P(\text{no 1 on any die}) = \left( \frac{5}{6} \right)^{100}$$

Step 2: Probability of at least one die showing a 1
The event $A$ represents the set of all outcomes where at least one die shows a 1. This is the complement of the event where none of the dice shows a 1. Therefore, the probability of $A$ (at least one die shows a 1) is:
$$P(A) = 1 - P(\text{no 1 on any die}) = 1 - \left( \frac{5}{6} \right)^{100}$$

Thus, the probability that at least one die shows a 1 is:
$$P(A) = 1 - \left( \frac{5}{6} \right)^{100}$$

Conclusion:
The correct answer is $\boxed{1 - \left(\frac{5}{6}\right)^{100}}$, which corresponds to Option C.