Respuesta :
Answer:
The probability of event A is [tex]\frac{7}{11}[/tex].
The probability of event B is [tex]\frac{6}{11}[/tex].
Step-by-step explanation:
The sample space of rolling a fair six-sided dice is as follows:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Now consider the experiment of the computing the sum of the two rolls as follows:
X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Number of total outcomes, N = 11.
The probability of an event E is the ratio of the number of favorable outcomes to the total number of outcomes.
[tex]P(E)=\frac{n(E)}{N}[/tex]
The event A is defined as the sum is greater than 5.
The sample space of A is:
A = {6, 7, 8, 9, 10, 11, 12}
n (A) = 7
Compute the probability of event A as follows:
[tex]P(A)=\frac{n(A)}{N}=\frac{7}{11}[/tex]
Thus, the probability of event A is [tex]\frac{7}{11}[/tex].
The event B is defined as the sum is an even number.
The sample space of B is:
B = {2, 4, 6, 8, 10, 12}
n (B) = 6
Compute the probability of event B as follows:
[tex]P(B)=\frac{n(B)}{N}=\frac{6}{11}[/tex]
Thus, the probability of event B is [tex]\frac{6}{11}[/tex].
