Probability Question from Strategy Guide 5 - Chapter 4- p 62

[skyjarrett12]

Q: Molly is rolling a number cube with faces numbered 1-6 repeatedly. WHen she receives a 1, she will stop rolling the cube. What is the probability that Molly will roll the die less than 4 times before stopping?

ANS. 1/6 + (5/6)(1/6) + (5/6)(5/6)(1/6) = 1/6 + 5/26 + 25/216 = 91/216

I'm trying to understand why it wouldn't be 1/6 + 1/6 + 1/6.

[kyle_proctor]

I would simply find the probability that she does NOT roll a "1" on the first three rolls. Compute this number and subtract from 1:

1 - [(5/6)(5/6)(5/6)] = 91/216

I will let the experts explain. I have a difficult time with probability. In the case above, we are assuming in her first three rolls she will not roll a "1".

Thus, probability equals 125/216

To find the probability the probability we are lookign for, simply subtract the above probability from one, yielding 91/216.

[skyjarrett12]

Kyle_proctor - thank you so much for your response. that helps in thinking through how to approach the problem.

Still hoping that an expert with help to explain why it wouldn't be 1/6+1/6+1/6 (i.e) P(she will on first) OR P(She wil on 2nd) OR P(she will on 3rd).

[tim]

P(2nd roll) is the probability that you DON'T roll a 1 on the 1st roll, because the only way to get to the 2nd roll is to avoid a 1 on the 1st roll. this is why we need to multiply the 1/6 by 5/6. does that help?