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 lett
er 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?
2 answers:
Answer:
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%
Answer:
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%
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