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mameadows
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you roll 2 dice. what is the probability that at least one will be a four?
 one year ago
 one year ago
mameadows Group Title
you roll 2 dice. what is the probability that at least one will be a four?
 one year ago
 one year ago

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apihliah Group TitleBest ResponseYou've already chosen the best response.0
11/36? just thinking
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
@mameadows Have you found the number of combinations where at least one die shows a four?
 one year ago

mameadows Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mameadows Group TitleBest ResponseYou've already chosen the best response.0
thanks :)
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
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: \[P(at\ least\ one\ 4)=\frac{5}{36}+\frac{5}{36}+\frac{1}{36}=\frac{11}{36}\]
 one year ago
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