Question

1. in a game of chance, 3 dice are thrown at the same time. what is the probability that all three will land on a 6?

1/6

1/18

1/30

1/216

Explanation:

Step1: Calculate the probability of one die landing on 6

The probability of a single die landing on 6 is \(P(\text{one }6)=\frac{1}{6}\) since there are 6 possible outcomes (1 - 6) and only 1 favorable outcome (6).

Step2: Use the multiplication rule for independent events

Since the throws of the three dice are independent events, the probability that all three dice land on 6 is \(P(\text{three }6s)=P(\text{one }6)\times P(\text{one }6)\times P(\text{one }6)\)
Substitute \(P(\text{one }6)=\frac{1}{6}\) into the formula: \(P(\text{three }6s)=\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}=\frac{1}{216}\)

Answer:

\(\frac{1}{216}\) (the fourth option)