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Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.
The probability that the sum of the numbers rolled is either even or a multiple of 5 is
The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is

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MathPhys

Answer:

11/18

5/9

Step-by-step explanation:

Make a table showing all the possible sums for two dice:

[tex]\left[\begin{array}{ccccccc}&1&2&3&4&5&6\\1&2&3&4&5&6&7\\2&3&4&5&6&7&8\\3&4&5&6&7&8&9\\4&5&6&7&8&9&10\\5&6&7&8&9&10&11\\6&7&8&9&10&11&12\end{array}\right][/tex]

Of the 36 possible sums, 18 are even, 7 are multiples of 5, and 3 are both even and multiple of 5.  So the probability is:

P(even or multiple of 5) = (18 + 7 − 3) / 36 = 22/36 = 11/18

Of the 36 possible sums, 12 are multiples of 3, 9 are multiples of 4, and 1 is a multiple of both 3 and 4.  So the probability is:

P(multiple of 3 or 4) = (12 + 9 − 1) / 36 = 20/36 = 5/9

Eduard22sly

1. The probability that the sum of the numbers rolled is either even or a multiple of 5 is 11/18

2. The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is 5/9

Possible outcomes

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1. How to determine the probability of even or a multiple of 5

  • Sample space (nS) = 36
  • Event of even numbers (nE) = 18
  • Event of multiply of 5 (n5) = 7
  • NOTE = 3 are both even and multiple of 5
  • Probability of even or a multiple of 5 [P(E or 5)] =?

P(E or 5) = P(E) + P(5) - P(3)

P(E or 5) = 18/36 + 7/36 - 3/36

P(E or 5) = (18 + 7 - 3) / 36

P(E or 5) = 22/36

P(E or 5) = 11/18

2. How to determine the probability of a multiple of 3 or 4

  • Sample space (nS) = 36
  • Event of multiply of 3 (n3) = 12
  • Event of multiply of 4 (n4) = 9
  • NOTE = 1 is both multiple of 3 and 4
  • Probability of multiple of 3 or 4 [P(3 or 4)] =?

P(3 or 4) = P(3) + P(4) - P(1)

P(3 or 4) = 12/36 + 9/36 - 1/36

P(3 or 4) = (12 + 9 - 1) / 36

P(3 or 4) = 20/36

P(3 or 4) = 5/9

Learn more about probability:

https://brainly.com/question/18878381

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