Solve 9(5-21/7)+4(2)
1 answer:
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Answer:
The probability of event <em>A</em> is .
The probability of event <em>B</em> is .
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, <em>N</em> = 11.
The probability of an event <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.
The event <em>A</em> is defined as the sum is greater than 5.
The sample space of <em>A</em> is:
A = {6, 7, 8, 9, 10, 11, 12}
n (A) = 7
Compute the probability of event <em>A</em> as follows:
Thus, the probability of event <em>A</em> is .
The event <em>B</em> is defined as the sum is an even number.
The sample space of <em>B</em> is:
B = {2, 4, 6, 8, 10, 12}
n (B) = 6
Compute the probability of event <em>B</em> as follows:
Thus, the probability of event <em>B</em> is .