a We have a jar with

5

red,

4

white, and

3

blue balls. We want to find the probability of choosing

3

white balls from the jar. This probability is the ratio of the number of ways to choose

3

white balls from the jar and the number of ways to choose

3

balls from the jar. The order of choosing the balls is not important, so each selection is a combination.

Combination
Selection of objects in which the order is not important.

First, let's find the number of ways in which we can select

3

white balls from

4

present in the jar. To do so, let's recall the formula for the number of combinations in which we choose

r

objects from

n

possible.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

We choose

3

white balls from

4

possible, so

r = 3

and

n = 4 .

Let's substitute these values into the above formula and simplify.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

SubstituteII

$n = 4$ , $r = 3$

_{4} C_{3} = \frac{4 !}{3 ! \cdot ( 4 - 3 ) !}

SubTerm

Subtract term

_{4} C_{3} = \frac{4 !}{3 ! \cdot 1 !}

$n! = n \cdot (n - 1)!$

_{4} C_{3} = \frac{4 \cdot 3 !}{3 ! \cdot 1 !}

ReduceFrac

$\frac{a}{b} = \frac{a / 3 !}{b / 3 !}$

_{4} C_{3} = \frac{4}{1 !}

$1! = 1$

_{4} C_{3} = \frac{4}{1}

CalcQuot

Calculate quotient

_{4} C_{3} = 4

We have

4

ways to choose the white balls. Now, since we want to find the probability of choosing

3

white balls, we should also find the number of ways of choosing

3

balls from the jar. Let's find the number of balls.

5 + 4 + 3 = 12

We have

12

balls from which we choose

3

balls. Let's again use the formula for the number of combinations

_{n} C_{r},

this time substituting

12

for

n

and

3

for

r .

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

SubstituteII

$n = 12$ , $r = 3$

_{12} C_{3} = \frac{1 2 !}{3 ! \cdot ( 1 2 - 3 ) !}

SubTerm

Subtract term

_{12} C_{3} = \frac{1 2 !}{3 ! \cdot 9 !}

▼

Simplify right-hand side

$n! = n \cdot (n - 1)!$

_{12} C_{3} = \frac{1 2 \cdot 1 1 !}{3 ! \cdot 9 !}

$n! = n \cdot (n - 1)!$

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0 !}{3 ! \cdot 9 !}

$n! = n \cdot (n - 1)!$

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0 \cdot 9 !}{3 \cdot 2 ! \cdot 9 !}

ReduceFrac

$\frac{a}{b} = \frac{a / 9 !}{b / 9 !}$

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0}{3 \cdot 2 !}

$2! = 2$

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0}{3 \cdot 2}

Multiply

_{12} C_{3} = \frac{1 3 2 0}{6}

CalcQuot

Calculate quotient

_{12} C_{3} = 220

We have

220

ways of choosing

3

balls from the jar. The probability

P

of choosing

3

white balls from the jar is the ratio of the number of ways to choose

3

white balls from the jar

(4)

and the number of ways to choose

3

balls from the jar

(220) .

P = \frac{4}{2 2 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

P = \frac{1}{5 5}

UseCalc

Use a calculator

P = 0.018181 \dots

We will write the above probability as a percent by multiplying its decimal value by

100 % .

P = 0.018181 \dots \cdot 100 % \approx 1.81 %

Exercise

This Solution is Missing. Mathvizy Can Solve This Problem For You