Question

in a certain board game, a 12 - sided number cube showing numbers 1 through 12 is rolled. in this game, a number cube must be rolled until a number 9 or higher appears. what is the probability that the first such number is on the 3rd roll?
\\( \frac { 1 } { 27 } \\)
\\( \frac { 4 } { 27 } \\)
\\( \frac { 8 } { 27 } \\)
\\( \frac { 19 } { 27 } \\)

Explanation:

Step1: Calculate the probability of rolling a number less than 9

Numbers less than 9 are 1 - 8. So the probability \(P(\text{less than }9)=\frac{8}{12}=\frac{2}{3}\).

Step2: Calculate the probability of rolling a number 9 or higher

Numbers 9 - 12. So the probability \(P(\text{9 or higher})=\frac{4}{12}=\frac{1}{3}\).

Step3: Calculate the probability that the first 9 or higher is on the 3rd roll

This means the first two rolls are less than 9 and the third roll is 9 or higher. Using the multiplication rule for independent events: \(P = (\frac{2}{3})\times(\frac{2}{3})\times(\frac{1}{3})\).

$$P=\frac{2\times2\times1}{3\times3\times3}=\frac{4}{27}$$

Answer:

\(\frac{4}{27}\) (the second option)