Question

A rack of 15 billiard balls is shown. If one ball is selected at​ random, determine the odds against it containing a number greater than or equal to 7.

Answer 1

Using the probability and odds concepts, it is found that the odds against it containing a number greater than or equal to 7 is $\frac{2}{3}$.

• A probability is the number of desired outcomes divided by the number of total outcomes.
• An odd is the number of desired outcomes divided by the number of non-desired outcomes.

In the rack, there are 15 balls, numbered from 1 to 15. Of those, 6 are less than 7(against it containing a number greater than or equal to 7 is equivalent to it containing a number less than 7), thus:

• There are 6 desired outcomes.
• There are 9 non-desired outcomes.

The odd is:

$\frac{6}{9} = \frac{2}{3}$

The odds against it containing a number greater than or equal to 7 is $\frac{2}{3}$.

A similar problem is given at brainly.com/question/21094006