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.