has critical points where
:
where is any integer. We get solutions in the interval
for
, for which
.
At this critical point, we have .
At the endpoints of the given interval, we have and
.
So we have the extreme values
Two number cubes are rolled for two separate events:
1 answer:
Answer:
<u>Step-by-step explanation:</u>
The probability of Event B given Event A = the intersection of Event A and B divided by the probability of Event A. (see below for the symbols)
P(A) = (1, 6), (1, 5), (1, 4), (1, 3), (1, 2), (1, 1)
(2, 6), (2, 5), (2, 4), (2, 3), (2, 2), (2, 1)
(3, 6), (3, 5), (3, 4), (3, 3), (3, 2), (3, 1)
(4, 5), (4, 4), (4, 3), (4, 2), (4, 1)
(5, 4), (5, 3), (5, 2), (5, 1)
(6, 3), (6, 2), (6, 1)
= 30
P(B) = (1, 2), (2, 1) sum = 3
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1) sum = 6
(3, 6), (4, 5), (5, 4), (5, 4), (6, 3) sum = 9
(6, 6) sum = 12
= 12
P(A ∩ B) = (1, 2), (2, 1)
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
(3, 6), (4, 5), (5, 4), (5, 4), (6, 3)
= 11
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