Finite Math Examples

$x = 3$ , $n = 12$ , $p = 0.46$

Step 1

Use the formula for the probability of a binomial distribution to solve the problem.

$p (x) = {}_{12}C_{3} \cdot p^{x} \cdot q^{n - x}$

Step 2

Find the value of ${}_{12}C_{3}$ .

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Step 2.1

Find the number of possible unordered combinations when $r$ items are selected from $n$ available items.

${}_{12}C_{3} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 2.2

Fill in the known values.

$\frac{(12)!}{(3)! (12 - 3)!}$

Step 2.3

Simplify.

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Step 2.3.1

Subtract $3$ from $12$ .

$\frac{(12)!}{(3)! (9)!}$

Step 2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9!}{(3)! (9)!}$

Step 2.3.3

Cancel the common factor of $9!$ .

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Step 2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9!}{(3)! (9)!}$

Step 2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10}{(3)!}$

Step 2.3.4

Simplify the numerator.

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Step 2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10}{(3)!}$

Step 2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320}{(3)!}$

Step 2.3.5

Simplify the denominator.

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Step 2.3.5.1

Expand $(3)!$ to $3 \cdot 2 \cdot 1$ .

$\frac{1320}{3 \cdot 2 \cdot 1}$

Step 2.3.5.2

Multiply $3 \cdot 2 \cdot 1$ .

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Step 2.3.5.2.1

Multiply $3$ by $2$ .

$\frac{1320}{6 \cdot 1}$

Step 2.3.5.2.2

Multiply $6$ by $1$ .

$\frac{1320}{6}$

Step 2.3.6

Divide $1320$ by $6$ .

$220$

Step 3

Fill the known values into the equation.

$220 \cdot {(0.46)}^{3} \cdot {(1 - 0.46)}^{12 - 3}$

Step 4

Simplify the result.

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Step 4.1

Raise $0.46$ to the power of $3$ .

$220 \cdot 0.097336 \cdot {(1 - 0.46)}^{12 - 3}$

Step 4.2

Multiply $220$ by $0.097336$ .

$21.41392 \cdot {(1 - 0.46)}^{12 - 3}$

Step 4.3

Subtract $0.46$ from $1$ .

$21.41392 \cdot {0.54}^{12 - 3}$

Step 4.4

Subtract $3$ from $12$ .

$21.41392 \cdot {0.54}^{9}$

Step 4.5

Raise $0.54$ to the power of $9$ .

$21.41392 \cdot 0.0039043$

Step 4.6

Multiply $21.41392$ by $0.0039043$ .

$0.08360649$