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Probability Problem

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Probability Problem

by Mr.Hollywood » Wed Feb 01, 2012 2:54 pm
Calvin is playing a game that requires him to roll a fair die repeatedly until she first rolls a 1, at which point he must stop rolling the die. What is the probability that Calvin will roll the die less than four times before stopping?

I know the answer is 1/6+(5/6x1/6)+(5/6x5/6x1/6)=91/216

Where I have problem is (5/6x1/6)and (5/6x5/6x1/6).
Can someone help to understand these two parts?

Thank you!
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by pemdas » Wed Feb 01, 2012 3:14 pm
consider first and foremust EVENT
How many Events and what? two Events - rolling 1 and not rolling 1
P(rolling 1)=1/6, P(not rolling 1)=1-1/6=5/6

prompt: What is the probability that Calvin will roll the die less than four times before stopping?
Answer: the required will be THREE rolls, or one roll=Success, rolling 1, i.e. 1/6
two events (independent events) 5/6 and 1/6, P(not rolling 1 and rolling 1)=(5/6)*(1/6)
three events (independent events) 5/6, 5/6 and 1/6, P(not rolling 1 and not rolling 1 and rolling 1)=(5/6)*(5/6)*(1/6)

add them up to obtain 1/6 + 5/36 + 25/216 = (36+30+25)/216 = 91/216

Mr.Hollywood wrote:Calvin is playing a game that requires him to roll a fair die repeatedly until she first rolls a 1, at which point he must stop rolling the die. What is the probability that Calvin will roll the die less than four times before stopping?

I know the answer is 1/6+(5/6x1/6)+(5/6x5/6x1/6)=91/216

Where I have problem is (5/6x1/6)and (5/6x5/6x1/6).
Can someone help to understand these two parts?

Thank you!
Success doesn't come overnight!
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