Answer to Question #326181 in Statistics and Probability for Robera

Question

A box of 80 candles consists of 30 defective and 50 non defective candles. If 10 of

this candles are selected at random by business man for sale, what is the probability

a. All will be defective

b. 6 will be non-defective

c. All will be non-defective

Answer 1

a. The total number of possible outcomes is the number of ways we can choose 10 candles of 80.

Favorable outcomes are when we choose 10 of 30 defective candles.

$P=\cfrac{\begin{pmatrix}30 \\10\end{pmatrix} }{\begin{pmatrix}80\\10\end{pmatrix}}=\cfrac{30!} {10!\cdot20!} \cdot\cfrac{10!\cdot70!} {80!} =0.0000182 .$

b. The total number of possible outcomes is the number of ways we can choose 10 candles of 80.

Favorable outcomes are when we choose 6 of 50 non-defective and 4 of 30 defective candles.

$P=\cfrac{\begin{pmatrix}50 \\6\end{pmatrix}\cdot\begin{pmatrix}30 \\4\end{pmatrix} }{\begin{pmatrix}80\\10\end{pmatrix}}=\cfrac{50!} {6!\cdot44!} \cdot\cfrac{30!} {4!\cdot26!} \cdot\cfrac{10!\cdot70!} {80!}=\\=0.2645 .$

c. The total number of possible outcomes is the number of ways we can choose 10 candles of 80.

Favorable outcomes are when we choose 10 of 50 non-defective candles.

$P=\cfrac{\begin{pmatrix}50 \\10\end{pmatrix} }{\begin{pmatrix}80\\10\end{pmatrix}}=\cfrac{50!} {10!\cdot40!} \cdot\cfrac{10!\cdot70!} {80!} =0.0062 .$