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For event A,
P(not A) = 1 - P(A)
For events A and B:
P(A or B) = P(A) + P(B) - P(A and B)
If the events are multually exclusive, the rule simplifies to
P(A or B) = P(A) + P(B)
For two independent events A and B,
P(A and B) = P(A) × P(B)
What is the probability, if you roll three dice, that at least one dice will be a 6?
The simplest approach is to use the negation rule with the multiplication rule. It is easy to calculate the probability for NOT getting a 6 in a roll, which is 5/6. And therefore The probability for NOT getting a 6 in three rolls is (5/6)³. And the 1 - (5/6)³ ≅ 0.421 or 42.1%.
A slightly more complicated approach is to use the additional rule and the multiplication rule iteratively. First we calculate the probability of rolling a six in either the first or the second dice:
P(A or B) = P(A) + P(B) - P(A and B) = 1/6 + 1/6 - (1/6)×(1/6) = 11/36
Then, we calculate again including the third dice:
P([A or B] or C) = P(A or B) + P(C) - P([A or B] and C)
= 11/36 + 1/6 - (11/36)×(1/6) ≅ 0.421 or 42.1%