Question

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

Answer 1

11/36? just thinking

Answer 2

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.

Answer 3

@mameadows Have you found the number of combinations where at least one die shows a four?

Answer 4

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!

Answer 5

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.

Answer 6

thanks :)

Answer 7

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:
\[P(at\ least\ one\ 4)=\frac{5}{36}+\frac{5}{36}+\frac{1}{36}=\frac{11}{36}\]