You have a rather strange die: Three faces are marked with the letter A, two faces with the letter B, and one face with the letter C. You roll the die until you get a B. What is the probability that a B does not appear during the first three rolls?

8/27 ≈ 29.6%

Step-by-step explanation:

Two of the six faces are B, which means four of the six faces are not B.

The probability of rolling not B three times is:

P = (4/6) ^3

P = (2/3) ^3

P = 8/27

P ≈ 29.6%

8/27 = 0.296 = 29.6%

Step-by-step explanation:

Given that the die faces are as follows:

A A A B B C

i. e:

P (rolls A) = 3/6

P (rolls B) = 2/6

P (rolls C) = 1/6

for any single roll,

P (rolls not B) = P (rolls A) + P (rolls C)

P (rolls not B) = 3/6 + 1/6 = 4/6 = 2/3

for 3 consecutive rolls

P (B does not appear) = P (rolls not B) * P (rolls not B) * P (rolls not B)

= P (rolls not B) ³

= (2/3) ³

= 8/27 = 0.296 = 29.6%