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11Alexandr11 [23.1K]
3 years ago
7

When rolling two fair, standard dice, what is the probability that the sum of the numbers rolled is a multiple of 3 or 4? Expres

s your answer as a common fraction.
Mathematics
2 answers:
harkovskaia [24]3 years ago
6 0
There are 6 times 6 = 36 possible outcomes when rolling two fair dice. Of those, 20 are multiplied of 3 or 4 as shown. Thus, the probability that the sum of the numbers rolled is a multiply of 3 or 4 is 20/36 = 5/9.
gizmo_the_mogwai [7]3 years ago
5 0
<span>There are 6 × 6 = 36 possible outcomes when rolling two fair dice. Of those, 20 are multiples of 3 or 4, as shown. Thus, the
probability that the sum of the numbers rolled is a multiple of 3 or 4 is 20/36 = 5/9.</span>
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Suppose that two teams are playing a series of games, each of which is independently won by team A with probability p and by tea
Lady_Fox [76]

Answer:

4 games played = 1/56

5 games played = 5/56

6 games played = 15/56

7 games played = 35/56

Step-by-step explanation:

The probability of each time winning is 0.5

So there are a couple of ways the series could go.

  1. Team A could win all first 4 matches. We can depict this as A-A-A-A
  2. Team A could win 4, while having lost 1.  Lets depict this a A-B-A-A-A
  3. Team A could win 4, while having lost 2. A-B-B-A-A-A
  4. Team A could win 4, while having lost 3. A-B-B-B-A-A-A

These 4 possibilities could be repeated with B winning as well. It should be noted that these are the only ways for the series to end. We find the number of permutations of each possibility above to find their probability.

I advise you to study 'how to permute identical objects' for this.

These permutations are stated below:

1. 4!/4! = 1

2. \frac{5!}{4! * 1!} = 5

3 \frac{6!}{4! * 2!} = 15

4 \frac{7!}{4! * 3!} = 35

These are they ways A or B could win. The total is 1+5+15+35 = 56. The answer given thus reflects the possibilities.

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4 years ago
A cube has a volume of 1000 what is the edge length of the cube
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A<span>≈</span><span>10
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