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After a late night of studying, Ebony decides to grab a latte before class so she can stay awake through her morning lecture. She has only a one-dollar bill, a five-dollar bill, and a ten-dollar bill in her wallet. She pulls one out and looks at it, but then she puts it back. Distracted by a flyer for a new campus organization, she randomly hands a bill from her wallet to the clerk. Construct a sample space, and find the probability that (a)Both bills have the same value. (b)The second bill is smaller than the first bill. (c)The value of each of the two bills is even. (d)The value of exactly two of the bills is odd. (e)The sum of the values of both bills is less than . Write your answers in exact, simplified form.

Respuesta :

MrRoyal

Answer:

[tex]P(Same\ Bill) = \frac{1}{3}[/tex]

[tex]P(Second<First\ Bill) = \frac{1}{3}[/tex]

[tex]P(Both\ Even) = \frac{1}{9}[/tex]

[tex]Pr(One\ Odd) = \frac{4}{9}[/tex]

[tex]P(Sum < 10) = \frac{1}{3}[/tex]

Step-by-step explanation:

Given

[tex]Bills: \$1, \$5, \$10[/tex]

[tex]Selection = 2\ bills[/tex]

The sample space is as follows:

This implies that we construct possible outcome that Ebony selects a bill, returns the bill and then select another.

This means that there are possibilities that the same bill is selected twice.

So, the sample space is as follows:

[tex]S = \{(1,1), (1,5), (1,10), (5,1), (5,5), (5,10), (10,1), (10,5), (10,10)\}[/tex]

[tex]n(S) = 9[/tex]

Solving (a): [tex]P(Same\ Bill)[/tex]

This means that the first and second bill selected are the same.

The outcome of this are:

[tex]Same = \{(1,1),(5,5),(10,10)\}[/tex]

[tex]n(Same\ Bill) = 3[/tex]

The probability is:

[tex]P(Same\ Bill) = \frac{n(Same\ Bill)}{n(S)}[/tex]

[tex]P(Same\ Bill) = \frac{3}{9}[/tex]

[tex]P(Same\ Bill) = \frac{1}{3}[/tex]

Solving (a): [tex]P(Second < First\ Bill)[/tex]

This means that the second bill selected is less than the first.

The outcome of this are:

[tex]Second < First = \{(1,5), (1,10), (5,10)\}[/tex]

[tex]n(Second < First) = 3[/tex]

The probability is:

[tex]P(Second<First\ Bill) = \frac{n(Second<First\ Bill)}{n(S)}[/tex]

[tex]P(Second<First\ Bill) = \frac{3}{9}[/tex]

[tex]P(Second<First\ Bill) = \frac{1}{3}[/tex]

Solving (c): [tex]P(Both\ Even)[/tex]

This means that the first and the second bill are even

The outcome of this are:

[tex]Both\ Even = \{(10,10)\}[/tex]

[tex]n(Both\ Even) = 1[/tex]

The probability is:

[tex]P(Both\ Even) = \frac{n(Both\ Even)}{n(S)}[/tex]

[tex]P(Both\ Even) = \frac{1}{9}[/tex]

Solving (e): [tex]P(Sum < 10)[/tex]

This question has missing details.

The correct question is to determine the probability that, the sum of both bills is less than 10

The outcome of this are:

[tex]One\ Odd = \{(1,10), (5,10), (10,1), (10,5)\}[/tex]

[tex]n(One\ Odd) = 4[/tex]

The probability is:

[tex]Pr(One\ Odd) = \frac{n(One\ Odd)}{n(S)}[/tex]

[tex]Pr(One\ Odd) = \frac{4}{9}[/tex]

 

Solving (d): [tex]P(One\ Odd)[/tex]

This question has missing details.

The correct question is to determine the probability that, exactly one of the bills is 0dd

The outcome of this are:

[tex]Sum < 10 = \{(1,1), (1,5), (5,1)\}[/tex]

[tex]n(Sum < 10) = 3[/tex]

The probability is:

[tex]P(Sum < 10) = \frac{n(Sum < 10)}{n(S)}[/tex]

[tex]P(Sum < 10) = \frac{3}{9}[/tex]

[tex]P(Sum < 10) = \frac{1}{3}[/tex]

 

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