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There are seven quarters in the bottom of a tote bag. Three of those quarters were minted in 2019, two were minted in 2001, and two were minted in 2008. What is the probability of selecting two quarters that were both minted in years other than 2019 if the first was not replaced before the second was selected?

Sagot :

MrRoyalidn

Answer:

[tex]P(Not\ 2019) = \frac{2}{7}[/tex]

Step-by-step explanation:

Given

[tex]n(2019)= 3[/tex]

[tex]n(2001)= 2\\[/tex]

[tex]n(2008)= 2[/tex]

[tex]n = 7[/tex] --- total

Required

[tex]P(Not\ 2019)[/tex]

When two quarters not minted in 2019 are selected, the sample space is:

[tex]S = \{(2001,2001),(2001,2008),(2008,2001),(2008,2008)\}[/tex]

So, the probability is:

[tex]P(Not\ 2019) = P(2001,2001)\ or\ P(2001,2008)\ or\ P(2008,2001)\ or\ P(2008,2008)[/tex]

[tex]P(Not\ 2019) = P(2001,2001) + P(2001,2008) + P(2008,2001) + P(2008,2008)[/tex]

[tex]P(2001,2001) = P(2001) * P(2001)[/tex]

Since it is a selection without replacement, we have:

[tex]P(2001,2001) = \frac{n(2001)}{n} * \frac{n(2001)-1}{n - 1}[/tex]

[tex]P(2001,2001) = \frac{2}{7} * \frac{2-1}{7 - 1}[/tex]

[tex]P(2001,2001) = \frac{2}{7} * \frac{1}{6}[/tex]

[tex]P(2001,2001) = \frac{1}{7} * \frac{1}{3}[/tex]

[tex]P(2001,2001) = \frac{1}{21}[/tex]

[tex]P(2001,2008) = P(2001) * P(2008)[/tex]

Since it is a selection without replacement, we have:

[tex]P(2001,2008) = \frac{n(2001)}{n} * \frac{n(2008)}{n - 1}[/tex]

[tex]P(2001,2008) = \frac{2}{7} * \frac{2}{7 - 1}[/tex]

[tex]P(2001,2008) = \frac{2}{7} * \frac{2}{6}[/tex]

[tex]P(2001,2008) = \frac{2}{7} * \frac{1}{3}[/tex]

[tex]P(2001,2008) = \frac{2}{21}[/tex]

[tex]P(2008,2001) = P(2008) * P(2001)[/tex]

Since it is a selection without replacement, we have:

[tex]P(2008,2001) = \frac{n(2008)}{n} * \frac{n(2001)}{n - 1}[/tex]

[tex]P(2008,2001) = \frac{2}{7} * \frac{2}{7 - 1}[/tex]

[tex]P(2008,2001) = \frac{2}{7} * \frac{2}{6}[/tex]

[tex]P(2008,2001) = \frac{2}{7} * \frac{1}{3}[/tex]

[tex]P(2008,2001) = \frac{2}{21}[/tex]

[tex]P(2008,2008) = P(2008) * P(2008)[/tex]

Since it is a selection without replacement, we have:

[tex]P(2008,2008) = \frac{n(2008)}{n} * \frac{n(2008)-1}{n - 1}[/tex]

[tex]P(2008,2008) = \frac{2}{7} * \frac{2-1}{7 - 1}[/tex]

[tex]P(2008,2008) = \frac{2}{7} * \frac{1}{6}[/tex]

[tex]P(2008,2008) = \frac{1}{7} * \frac{1}{3}[/tex]

[tex]P(2008,2008) = \frac{1}{21}[/tex]

So:

[tex]P(Not\ 2019) = P(2001,2001) + P(2001,2008) + P(2008,2001) + P(2008,2008)[/tex]

[tex]P(Not\ 2019) = \frac{1}{21} + \frac{2}{21} +\frac{2}{21} +\frac{1}{21}[/tex]

Take LCM

[tex]P(Not\ 2019) = \frac{1+2+2+1}{21}[/tex]

[tex]P(Not\ 2019) = \frac{6}{21}[/tex]

Simplify

[tex]P(Not\ 2019) = \frac{2}{7}[/tex]

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