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Carlos has chosen 12 different CDs he would like to buy: 4 are rap music, 5 are country, and 3 are heavy metal. He has only enough money to buy 5 of them (each CD costs the same price). So he selects 5 of them at random. What is the probability that his purchase includes at least one CD from each of the three genres.

Sagot :

Answer:

The probability is 0.7449

Step-by-step explanation:

Given

[tex]n = 12[/tex] ---- total

[tex]r = 5[/tex] --- selection

[tex]Genre =\{Rap(4), Country (5), Heavy\ metal (3)\}[/tex]

Required

Probability of buying at least 1 of each genre

First, we calculate the total possible selection.

To select 5 CDs from a total of 12, we use:

[tex]^nC_r = \frac{n!}{(n - r)!r!}[/tex]

[tex]^{12}C_5 = \frac{12!}{(12 - 5)!5!}[/tex]

[tex]^{12}C_5 = \frac{12!}{7!5!}[/tex]

Expand

[tex]^{12}C_5 = \frac{12*11*10*9*8*7!}{7!*5*4*3*2*1}[/tex]

[tex]^{12}C_5 = \frac{12*11*10*9*8}{5*4*3*2*1}[/tex]

[tex]^{12}C_5 = \frac{95040}{120}[/tex]

[tex]^{12}C_5 = 792[/tex]

So, the total selection is:

[tex]Total = 792[/tex]

To select at least 1 from each genre, there are 6 possible scenarios.

And they are:

[tex]\begin{array}{ccc}{Heavy\ Metal (3)} & {Rap (4)} & {Country(5)} & {3} & {1} & {1} & 2 & {1} & {2} & {2} & {2} & {1}& {1} & {2} & {2}& {1} & {3} & {1}& {1} & {1} & {3} \ \end{array}[/tex]

The possible selections for the given scenario is:

[tex]Possible = ^3C_3* ^4C_1 * ^5C_1 +^3C_2* ^4C_1 * ^5C_2 +^3C_2* ^4C_2 * ^5C_1 +^3C_1* ^4C_2 * ^5C_2 +^3C_1* ^4C_3 * ^5C_1 +^3C_1* ^4C_1 * ^5C_3[/tex]

Using a calculator, we have:

[tex]Possible = 1*4*5 +3*4*10 +3*6*5 +3*6*10+3*4*5+3*4*10[/tex]

[tex]Possible= 20 +120 +90 +180+60+120[/tex]

[tex]Possible = 590[/tex]

The probability is then calculated using:

[tex]Pr = \frac{Possible}{Total}[/tex]

[tex]Pr = \frac{590}{792}[/tex]

[tex]Pr = 0.7449[/tex]

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