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Mekhanik [1.2K]
2 years ago
15

Of three dice, two are labeled normally: 1, 2, 3, 4, 5 , and 6. The third die has sides labeled 2,4,6,8,10, and 12. Two dice are

rolled and their values are added. Find the ratio of the frequency of primes rolled using the first two ( ordinary) dice to the frequency of primes rolled using the unusual die and one ordinary die .
Mathematics
1 answer:
jolli1 [7]2 years ago
7 0

The ratio of the frequency of a prime rolled using the first two (ordinary)dice to the frequency of a prime rolled using the unusual die and one ordinary die is

  • \frac{15}{13}

<h3>The Cases and there Probable Results</h3>

There are a total of 36 equally probable ways the dice can fall

In the first case, primes are achieved for the following results:

  • 2 (one way possible),
  • 3 (two ways possible),
  • 5 (4 ways possible),
  • 7 (six ways possible), and
  • 11 (2 ways possible.

So primes are achieved in a total of 15 ways.

In the second case, the prime possibilities are

  • 3 (one way possible),
  • 5 (two ways),
  • 7 (3 ways),
  • 11 (3 ways),
  • 13 (3 ways), and
  • 17 (just one way).

The total is 13 ways.

<h3>How to calculate the ratio of the frequency</h3>

Therefore

How to calculate the ratio of the frequency is Mathematically given as

Frequency ratio is \frac{15}{13}

For more information on dice,

visithttps://brainly.com/question/2264527

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