Description
Need assistance in completing only 5 problems for Finite math subject.
Attached is the document to view the world problems, please be sure to show all work and steps, and not just the answers.
If everything goes well I will be happy to work with you again.
Due date is 10/01/2017 before 12:00AM
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Explanation & Answer
The solutions are ready. Please ask if something is unclear.docx and pdf files are identical
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1.
a. This may occur in three ways:
1) the first is red, the second is red, the third is white;
2) the first is red, the second is white, the third is red;
3) the first is white, the second is red, the third is red.
These events are incompatible so their probabilities add up. For 1), the probability is
3⁄ ∙ 2⁄ ∙ 2⁄ = 1⁄ . For 2), 3⁄ ∙ 2⁄ ∙ 2⁄ = 1⁄
3
2
2
1
8
7
6
28
8
7
6
28 and for 3) is ⁄8 ∙ ⁄7 ∙ ⁄6 = ⁄28.
So the probability is 𝟑⁄𝟐𝟖.
b. There are 6 ways: 1) red, white, blue; 2) red, blue, white; 3) blue, red, white; 4) blue,
white, red; 5) white, red, blue; 6) white, blue, red.
The probability for 1) is 3⁄8 ∙ 2⁄7 ∙ 3⁄6 = 3⁄56. All other are the same (8 ∙ 7 ∙ 6 in the
denominator and 3 ∙ 2 ∙ 3 in different orders at the numerator). So the probability is
18⁄ = 𝟗⁄ .
𝟐𝟖
56
c. None blue means we draw red or white marble each time. For the first draw, the
probability of not blue is 5⁄8. For the second draw the probability of not blue (provided
the first is not blue) is 4⁄7 and for the third 3⁄6 = 1⁄2.
The probability of all three are not blue is 5⁄8 ∙ 4⁄7 ∙ 1⁄2 = 𝟓⁄𝟐𝟖.
d. 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑏𝑙𝑢𝑒) = 1 − 𝑃(𝑛𝑜𝑛𝑒 𝑏𝑙𝑢𝑒) = �...
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