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Answer:
(a) The Probability that at most three lines are in use is 0.70.
(b) The Probability that fewer than three lines are in use is 0.45.
(c) The Probability that at least three lines are in use is 0.55.
(d) The Probability that between two and five lines, inclusive, are in use is 0.70.
(e) The Probability that between two and four lines, inclusive, are not in use is 0.35.
(f) The Probability that at least four lines are not in use is 0.70.
Step-by-step explanation:
<u>The complete question is: </u>A mail-order company business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table.
X 0 1 2 3 4 5 6
P(X) 0.10 0.15 0.20 0.25 0.20 0.05 0.05
Calculate the probability of each of the following events.
(a) {at most three lines are in use}
(b) {fewer than three lines are in use}
(c) {at least three lines are in use}
(d) {between two and five lines, inclusive, are in use}
(e) {between two and four lines, inclusive, are not in use}
(f) {at least four lines are not in use}
Now considering the above probability distribution;
(a) The Probability that at most three lines are in use is given by = P(X 3)
P(X 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.10 + 0.15 + 0.20 + 0.25
= 0.70
(b) The Probability that fewer than three lines are in use is given by = P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= 0.10 + 0.15 + 0.20
= 0.45
(c) The Probability that at least three lines are in use is given by = P(X 3)
P(X 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
= 0.25 + 0.20 + 0.05 + 0.05
= 0.55
(d) The Probability that between two and five lines, inclusive, are in use is given by = P(2 X
5)
P(2 X
5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 0.20 + 0.25 + 0.20 + 0.05
= 0.70
(e) The Probability that between two and four lines, inclusive, are not in use is given by = 1 - P(2 X
4)
1 - P(2 X
4) = 1 - [P(X = 2) + P(X = 3) + P(X = 4)]
= 1 - [0.20 + 0.25 + 0.20]
= 1 - 0.65 = 0.35
(f) The Probability that at least four lines are not in use is given by = 1 - Probability that at least four lines are in use = 1 - P(X 4)
1 - P(X 4) = 1 - [P(X = 4) + P(X = 5) + P(X = 6)]
= 1 - [0.20 + 0.05 + 0.05]
= 1 - 0.30 = 0.70
Explanation : If 66 is a factor of this unknown value, then 22 must be as well considering that 22 is a factor off 66. Let's say that this large value is 330. It is a multiple of 66, as 66 5 = 330. At the same time 22
15 = 330, so 330 is a multiple of 22 as well - or vice versa, 12 is a factor of 330.
We can also tell that 15, 22 fit into 330 through another approach. 22 3 = 66, and 66
5 = 330, so 5
3 = 15 - the same value. This proves that 22 will always be a factor of a value that is the factor of 66.