You can put this solution on YOUR website! a) One die can be 5 or the other, or both.
This probability is the same as 1 - Pr{no fives are rolled} which is easier to calculate:
Pr{no fives are rolled} = (5/6)*(5/6) = 25/36
So Pr{ 1 or more 5's are rolled } = 1 - 25/36 = (36-25)/36 =
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b) By inspection, Pr { 1 or more 2's are rolled } =
The reason is we've just substituted '2' for '5' and either number has the same probability of being rolled.
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c) Pr { at least one 2 or 5 is rolled } = 1 - Pr { neither a 2 nor a 5 is rolled }
= 1 - (4/6)*(4/6) = 1 - 16/36 = (36 - 16)/36 = 20/36 = 5/9
Pr { at least one 2 or 5 is rolled } =
What may be a little surprising is we're only interested in two of the six numbers on each die but the probability of getting one or more of those two numbers (with two dice) is more than 50%.
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NOTE: You _can_ use the first two probabilities to solve part (c). However, one must be very careful when doing so because you must subtract any outcomes that are over counted. One must be careful not to just do something like this (11/36) + (11/36) = 22/36 (which is incorrect). It is incorrect because we've counted the cases (2 then 5) and (5 then 2) twice, and therefore we must subtract 2/36 from the 22/36, bringing us into agreement with the answer found in part (c). The cases (2 then 5) and (5 then 2) are the intersection cases between the set of outcomes in (a) and (b).
