Question

Howard rolls a fair, 6 sided die 6 times. He is recording how many times he rolls a multiple of 3. What is the probability of Howard rolling exactly 3 multiples of 3? Show your work.

Answer #1

If a 6 sided die is rolled, then in 2 ways it can be multiple of 3, one is if 3 is rolled, another is if 6 is rolles. And in a die total number of possible outcome is 6. So if getting a number multiple of 3 is defined as success, then success probability, p =2/6 =1/3

Failure probability, q=2/3

Total number of trial, n=6

Number of success=3

Let Number of times Howard gets a amultiple of 3 is denoted by a random variable X.

X~ Bin(6, 1/2)

So, P(X=3)=
(2/3)^(6-3)*(1/3)^3= **0.2195**

The probability that Howard gets multiple of 3 exactly 3 times is 0.2195.

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