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you roll 2 dice. what is the probability that at least one will be a four?

Mathematics
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11/36? just thinking
The sample space has 36 possible combinations of numbers. These can be set out in column form as follows: 6,6 5,6 4,6 3,6 2,6 1,6 6,5 5,5 4,5 3,5 2,5 1,5 6,4 5,4 4,4 3,4 2,4 1,4 6,3 5,3 4,3 3,3 2,3 1,3 6,2 5,2 4,2 3,2 2,2 1,2 6,1 5,1 4,1 3,1 2,1 1,1 You need to count the number of outcomes where at least one die shows a four and divide the result by 36 to find the required probability.
@mameadows Have you found the number of combinations where at least one die shows a four?

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well, I see from your answer that there are 11 possible combinations, but is there a formula that I can use to make this quicker to solve? so I don't always have to draw out all of the possible outcomes? thank you by the way... your answer really helped!
No doubt there is an algebraic way to establish the number of combinations containing at least one four. However I think that making a table is the clearest way to solve this question.
thanks :)
Here is another way to solve this question: Let event A be getting 4 on die #1 and 1, 2, 3, 5 or 6 on die #2. The probability of getting 4 on die #1 is 1/6 and the probability of getting 1, 2, 3, 5 or 6 on die #2 is 5/6. Therefore the probability of event A is \[P(A)=\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}\] Let event B be getting 4 on die #2 and 1, 2, 3, 5 or 6 on die #1. The probability of getting 4 on die #2 is 1/6 and the probability of getting 1, 2, 3, 5 or 6 on die #1 is 5/6. Therefore the probability of event B is \[P(B)=\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}\] Let event C be getting 4 on die #1 and 4 on die #2. The probability of event C is \[P(C)=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\] Events A, B and C are mutually exclusive. Therefore the probability of getting at least one four is given by their sum, as follows: \[P(at\ least\ one\ 4)=\frac{5}{36}+\frac{5}{36}+\frac{1}{36}=\frac{11}{36}\]

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