Question

in a certain board game, a 12 - sided number cube showing numbers 1 - 12 is rolled. if three such number cubes are rolled, what is the probability that at least 1 of them shows a 2?
options: 0.0006, 0.2297, 0.7703, 0.9994

Explanation:

Step1: Find probability of not getting 2

A 12 - sided cube has numbers 1 - 12. The probability of not getting a 2 on one roll is $P(\text{not }2)=\frac{11}{12}$.

Step2: Probability of not getting 2 in three rolls

Since the rolls are independent, the probability of not getting a 2 in three rolls is $(\frac{11}{12})^3$. Calculate $(\frac{11}{12})^3=\frac{11\times11\times11}{12\times12\times12}=\frac{1331}{1728}\approx0.7703$.

Step3: Probability of at least one 2

The probability of at least one 2 is the complement of getting no 2s. So $P(\text{at least one }2)=1 - P(\text{no }2s)=1 - 0.7703 = 0.2297$.

Answer:

0.2297