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navik [9.2K]
3 years ago
12

A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3

, 4, and 5) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 5, and another is 213. (Enter your answers as a comma-separated list. Enter Ø for the empty set.)(a) List the outcomes in SS={______________}b) Let A denote the event that exactly one book must be examined. Whatoutcomes are in A?A = {______________}(c) Let B be the event that book 5 is the one selected. What outcomes arein B?B= {____________________}(d) Let C be the event that book 1 is not examined. What outcomes are inC?C ={_________________________}
Mathematics
1 answer:
Ainat [17]3 years ago
7 0

Answer:

1. Outcome of SS = {3,4,5,13,14,15,23,24,25,123,124,125,213,214,215}

2. Outcome of A = {3,4,5}

3. Outcome of B = {5,15,25,125,215}

4. Outcome of C = {3,4,5,23,24,25}

Step-by-step explanation:

1.

Let SS = Sample Space = Total possible events i.e. Event that a student stopped after picking a book from second printing (with no condition attached)

This means that there are tendencies that the student

1. Picks directly from the second prints: 3,4,5

2. Picks 1 book from first prints then second prints; 13,14,15,23,24,25

3. Picks 2 books from first prints the second prints : 123,124,125,213,214,215

Total number of possible outcomes = 15

Outcomes = {3,4,5,13,14,15,23,24,25,123,124,125,213,214,215}

2.

Let A = Event that exactly one book must be examined

This means that there are tendencies that the student picks directly from the second prints: 3,4,5

Total number of possible outcomes = 3

Outcome = {3,4,5}

3.

Let B = the event that book 5 is the one selected

This means that, the student didn't pick book 3 and 4.

Total number of outcomes = 5

Outcomes = {5,15,25,125,215}

4.

Let C = the event that book 1 is not examined

This means that the student didn't pick book 1

Total number of possible outcomes = 6

Outcome = {3,4,5,23,24,25}

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<img src="https://tex.z-dn.net/?f=%284v%5E%7B-3%7D%20%29%5E3%2F3v%5E%7B-8%7D" id="TexFormula1" title="(4v^{-3} )^3/3v^{-8}" alt=
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One is given the following equation:

\frac{(4v^-^3)^3}{3v^-^8}

Simplify the numerator, remember to raise every number inside the parenthesis to the exponent outside of the parenthesis. Bear in mind, an exponent raised to another exponent is equal to the exponent times the exponent it is raised to. Then simplify by multiplying the number by itself the number of times that the exponent indicates.

\frac{(4v^-^3)^3}{3v^-^8}

=\frac{4^3(v^-^3)^3}{3v^-^8}

=\frac{4^3v^-^9}{3v^-^8}

=\frac{64v^-^9}{3v^-^8}

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=\frac{64v^-^9}{3v^-^8}

=\frac{(64v^-^9)(v^(^-^1^)^(^-^8^))}{3}

Simplify, remember, multiplying two numbers with the same base that have an exponent is the same as adding the two exponents,

=\frac{(64v^-^9)(v^(^-^1^)^(^-^8^))}{3}

=\frac{(64v^-^9)(v^8)}{3}

=\frac{64v^-^9^+^8)}{3}

=\frac{64v^-^1}{3}

Now bring the variable to the denominator so that there are no negative exponents. Use a similar technique that was used to bring variables with exponents to the numerator.

=\frac{64v^-^1}{3}

=\frac{64}{3(v^(^-^1^)^(^-^1^))}

=\frac{64}{3v^1}

=\frac{64}{3v}

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