Finite Math Examples

$x = 3$ , $n = 5$ , $p = 0.3$

Step 1

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

$p (x) = {}_{5}C_{3} \cdot p^{x} \cdot q^{n - x}$

Step 2

Find the value of ${}_{5}C_{3}$ .

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Step 2.1

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

${}_{5}C_{3} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 2.2

Fill in the known values.

$\frac{(5)!}{(3)! (5 - 3)!}$

Step 2.3

Simplify.

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Step 2.3.1

Subtract $3$ from $5$ .

$\frac{(5)!}{(3)! (2)!}$

Step 2.3.2

Rewrite $(5)!$ as $5 \cdot 4 \cdot 3!$ .

$\frac{5 \cdot 4 \cdot 3!}{(3)! (2)!}$

Step 2.3.3

Reduce the expression by cancelling the common factors.

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Step 2.3.3.1

Cancel the common factor of $3!$ .

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Step 2.3.3.1.1

Cancel the common factor.

$\frac{5 \cdot 4 \cdot 3!}{(3)! (2)!}$

Step 2.3.3.1.2

Rewrite the expression.

$\frac{5 \cdot 4}{(2)!}$

Step 2.3.3.2

Multiply $5$ by $4$ .

$\frac{20}{(2)!}$

Step 2.3.4

Simplify the denominator.

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Step 2.3.4.1

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

$\frac{20}{2 \cdot 1}$

Step 2.3.4.2

Multiply $2$ by $1$ .

$\frac{20}{2}$

Step 2.3.5

Divide $20$ by $2$ .

$10$

Step 3

Fill the known values into the equation.

$10 \cdot {(0.3)}^{3} \cdot {(1 - 0.3)}^{5 - 3}$

Step 4

Simplify the result.

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Step 4.1

Raise $0.3$ to the power of $3$ .

$10 \cdot 0.027 \cdot {(1 - 0.3)}^{5 - 3}$

Step 4.2

Multiply $10$ by $0.027$ .

$0.27 \cdot {(1 - 0.3)}^{5 - 3}$

Step 4.3

Subtract $0.3$ from $1$ .

$0.27 \cdot {0.7}^{5 - 3}$

Step 4.4

Subtract $3$ from $5$ .

$0.27 \cdot {0.7}^{2}$

Step 4.5

Raise $0.7$ to the power of $2$ .

$0.27 \cdot 0.49$

Step 4.6

Multiply $0.27$ by $0.49$ .

$0.1323$