Respuesta :
Answer: The required probabilities are
(1) [tex]\dfrac{11}{18}.[/tex]
(2) [tex]\dfrac{5}{9}.[/tex]
Step-by-step explanation: Given that Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.
Let 'S' be the sample space for the experiment.
Then, S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), . . ., (6, 5), (6,6) }
That is, n(S) = 36.
(1) We are to find the probability that the sum of the numbers rolled is either even or a multiple of 5.
Let 'A' and 'B' be the events that the sum of the numbers rolled is an even number and a multiple of 5 respectively.
Then,
A = {(1,1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4,2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)},
and
B = {(1,4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}.
So, the event that the sum rolled is an even number or a multiple of 5 is given by
A ∪ B = {(1,1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4,2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6), (1,4), (2, 3), (3, 2), (4, 1)}
⇒ n(A ∪ B) = 22.
Therefore, the probability that the sum of the numbers rolled is either even or a multiple of 5 will be
[tex]P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}=\dfrac{22}{36}=\dfrac{11}{18}.[/tex]
(2) We are to find the probability that the sum of the numbers rolled is either a multiple of 3 or a multiple of 4.
Let 'C' and 'D' be the events that the sum of the numbers rolled is a multiple of 3 and a multiple of 4 respectively.
Then,
C = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)},
and
D = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}.
So, the event that the sum rolled is a multiple of 3 or a multiple of 4 is given by
C ∪ D = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6), (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2)}
⇒ n(C ∪ D) = 20.
Therefore, the probability that the sum of the numbers rolled is either a multiple of 3 or a multiple of 4 will be
[tex]P(C\cup D)=\dfrac{n(C\cup D)}{n(S)}=\dfrac{20}{36}=\dfrac{5}{9}.[/tex]
Thus, the required probabilities are
(1) [tex]\dfrac{11}{18}.[/tex]
(2) [tex]\dfrac{5}{9}.[/tex]
The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is [tex]\dfrac{10}{18}[/tex].
What is Probability?
The probability helps us to know the chances of an event occurring.
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
Given to us
- Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.
A table is drawn below showing the sum of the numbers that occur for the two dices.
A.) The probability that the sum of the numbers rolled is either even or a multiple of 5 is ___.
- Number of cases when an even number can occur = 18 (colored in green)
- Number of cases when a multiple of 5 can occur = 4 (colored in orange)
{Number 10 is not included as it is already included when we counted even number cases.}
- Total number of cases = n² = 6² = 36
Probability( either even or a multiple of 5)
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
[tex]\rm Probability=\dfrac{\text{ cases when the sum of the numbers rolled is either even or a multiple of 5 }}{\text{Total number of cases}}[/tex]
[tex]Probability = \dfrac{18+4}{36}\\\\Probability = \dfrac{22}{36}\\\\Probability = \dfrac{11}{18}\\[/tex]
Hence, the probability that the sum of the numbers rolled is either even or a multiple of 5 is [tex]\dfrac{11}{18}[/tex].
B.) The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is ___.
- Number of cases when a multiple of 3 can occur = 12(colored in purple)
- Number of cases when a multiple of 5 can occur = 8 (colored in orange)
{Number 12 is not included as it is already included when we counted multiple of 3.}
- Total number of cases = n² = 6² = 36
Probability(either a multiple of 3 or 4 )
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
[tex]\rm Probability=\dfrac{\text{ cases when the sum of the numbers rolled is either a multiple of 3 or 4 }}{\text{Total number of cases}}[/tex]
[tex]Probability = \dfrac{12+8}{36}\\\\Probability = \dfrac{20}{36}\\\\Probability = \dfrac{10}{18}\\[/tex]
Hence, the probability that the sum of the numbers rolled is either a multiple of 3 or 4 is [tex]\dfrac{10}{18}[/tex].
Learn more about Probability:
https://brainly.com/question/795909
