Question

A single shelf holds 27 books. On one particular shelf, 8 books are best sellers. If you choose to read 3 of them randomly, what is the probability all 3 chosen are best sellers? Give your answer as an exact fraction and reduce the fraction as much as possible.

Answer 1

Answer:

The probability of all chosen be the best:

                                             $\frac{\frac{8\cdot 7\cdot 6}{3!}}{\frac{27\cdot 26\cdot 25}{3!}}=\frac{8\cdot 7\cdot 6}{27\cdot 26\cdot 25}$

                                                        $=\frac{56}{2952}$

To determine:

Step-by-step explanation:

If you choose to read 3 of them randomly, what is the probability all 3 chosen are best sellers?

Information Fetching and Solution Steps:

• A single shelf holds 27 books.

• On one particular shelf, 8 books are best sellers.
• We have to choose to read 3 of them randomly

The formula to find the number of possible combinations

of 3 books from 27 books where order doesn't matter

                                 $C\left(n,k\right)=\frac{n!}{k!\left(n-k\right)!}$

So,

                                  $\:\:C\left(27,\:3\right)=\frac{27!}{3!\left(27-3\right)!}$

                                                  $=\frac{27!}{3!24!}$

                                                  $=\frac{27\cdot 26\cdot 25}{3!}$

And

The formula for the number of combinations of 3 vest sellers from 8 books will be:

                                  $C\left(8,\:3\right)=\frac{8!}{3!\left(8-3\right)!}$

                                                $=\frac{8!}{3!5!}$

                                                $=\frac{8\cdot 7\cdot 6}{3!}$

So, the probability of all chosen be the best:

                                             $\frac{\frac{8\cdot 7\cdot 6}{3!}}{\frac{27\cdot 26\cdot 25}{3!}}=\frac{8\cdot 7\cdot 6}{27\cdot 26\cdot 25}$

                                                        $=\frac{56}{2952}$

Keywords: probability, combination

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Answer 2

Answer:

$\large\boxed{\large\boxed{{\text{Probability all 3 chosen are best sellers}=56/2,925}}}$

Explanation:

1. Number of possible combinations

The number of possible combinations of 3 books from 27 books, whre the order does not matter, is given by the combinatory formula:

         $C(n,k)=\frac{n!}{k!(n-k)!}$

        $C(27,3)=\frac{27!}{3!(27-3)!}=\frac{27!}{3!24!}=\frac{27\cdot 26\cdot 25}{3!}$

2. Number of combinations of 3 bestsellers from 8 books that are best sellers

         $C(8,3)=\frac{8!}{3!(8-3)!}=\frac{8!}{3!5!}=\frac{8\cdot 7\cdot 6}{3!}$

3. Probability of all chosen are best sellers

   $\text{Probability of all chosen are best sellers}=\frac{\text{# combinations of 3 best sellers}}{\text{# total combinations}}$

   $\text{Probability of all chosen are best sellers}=\frac{\frac{8\cdot 7\cdot 6}{3!}}{\frac{27\cdot 26\cdot 25}{3!}}=\frac{8\cdot 7\cdot 6}{27\cdot 26\cdot 25}$

    $\text{Probability of all chosen are best sellers}=\frac{336}{17,550}\\\\ \text{Probability of all chosen are best sellers}=\frac{56}{2,925}$