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mjschlotterbeck
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When to use 1-X: Guide 5, 5th ed., Chapter 7, Problem 7

by mjschlotterbeck Sun Mar 03, 2013 3:43 pm

For one roll of a certain die, he probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
A) (1/6)^4
B)2(1/6)^3 + (1/6)^4
C)3(1/6)^3 x (5/6) + (1/6)^4
D)4(1/6)^3 x (5/6) + (1/6)^4
E)6(1/6)^3 x (5/6) + (1/6)^4

The answer says that you cannot use the 1-X trick easily and therefore have to calculate the probability of getting a two on all four rolls and the probability of rolling 3-twos and 1-non-two (four different combos: 2-2-2-x, 2-2-x-2, 2-x-2-2, x-2-2-2) and add their respective probabilities together. This leads to D)4(1/6)^3 x (5/6) + (1/6)^4.

Can you please elaborate on why you cannot use the 1-X rule easily here? I am trying to understand the nuances between 1-X and when I can and cannot use it.

For sake of trying, I believe you cannot use 1-X here because
1) You have to roll the die 4 times no matter what
2) The winning scenario involves 3 or 4 desired outcomes, opposed to 1
3) The 1-X seems to be employed when the person stops rolling after achieving the 1 desired outcome.
RonPurewal
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Re: When to use 1-X: Guide 5, 5th ed., Chapter 7, Problem 7

by RonPurewal Wed Mar 06, 2013 9:59 am

this is one of those things that people like to overcomplicate, but in reality it's not complicated.

basically, the whole "1 - whatever" thing involves calculating the probability that your desired event doesn't happen, and then subtracting that probability from 1 (to find the probability that it does happen).

so, the only rule you need is this:
if it's easier to calculate the probability that your event DOESN'T happen than to calculatethe probability that it DOES happen, then use the "1 - whatever" thing.

in this case, there are only two ways in which your event can happen:
* 3 twos and 1 non-two
* 4 twos

on the other hand, if you want the ways in which it doesn't happen, there are more of those:
* 2 twos and 2 non-twos
* 1 two and 3 non-twos
* 4 non-twos

so, yeah, the "1 - whatever" would be a bad move here, because you would have to do a longer calculation before subtracting from 1.