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8. There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random without replacement.
a. How many different ways are there of selecting the three balls?
b. What is the probability that one of the balls selected is the number 5?

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Answer:

a. 720

b. 0.3

Step-by-step explanation:

a.

Three balls are selected at random without replacement.

First ball can be selected in 10 ways

Second ball can be selected in 9 ways

Third ball can be selected in 8 ways

Thus, there are 10×9×8 =720 different ways of selecting three balls.

b.

There are three cases that satisfy this condition

  1. first ball picked is 5, the others are not
  2. first ball is not 5, second is 5, third is not
  3. first and second ball is not five, third ball is 5

Each probabilities are:

  1. [tex]\frac{1}{10} * \frac{9}{9} *\frac{8}{8} =\frac{1}{10}[/tex]
  2. [tex]\frac{9}{10} *\frac{1}{9} *\frac{8}{8} =\frac{1}{10}[/tex]
  3. [tex]\frac{9}{10} *\frac{8}{9} *\frac{1}{8} =\frac{1}{10}[/tex]

Thus, one of the balls selected is the number 5 is then [tex]\frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}[/tex]

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