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mameadows

  • 2 years ago

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

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  1. apihliah
    • 2 years ago
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    11/36? just thinking

  2. kropot72
    • 2 years ago
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    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.

  3. kropot72
    • 2 years ago
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    @mameadows Have you found the number of combinations where at least one die shows a four?

  4. mameadows
    • 2 years ago
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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!

  5. kropot72
    • 2 years ago
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    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.

  6. mameadows
    • 2 years ago
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    thanks :)

  7. kropot72
    • 2 years ago
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    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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