Question

There are 10 balls in a bag numbered from 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?
show work step by step

Answer 1

a) There are 720 different ways of selecting the three balls.

b) 30% probability that one of the balls selected is the number 5

Explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, the order in which the balls are selected is important. That is, (1,2,3) is a different outcome than (1,3,2). So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

`P_((n,x)) = (n!)/((n-x)!)`

a. How many different ways are there of selecting the three balls?

3 balls from a set of 10. So

`P_((10,3)) = (10!)/((10-3)!) = 720`

There are 720 different ways of selecting the three balls.

b. What is the probability that one of the balls selected is the number 5?

Total outcomes:

As found in a), 720.

Desired outcomes:

One of three balls being a 5. So 1 from a set of 3.

The other 2 being chosen from a set of 9. So

`P_((9,2))*P_((3,1)) = (9!)/((9-2)!)*(3!)/((3-1)!) = 216`

Probability:

`p = (D)/(T) = (216)/(720) = 0.3`

30% probability that one of the balls selected is the number 5