Finite Math Examples

$x = 1$ , $n = 3$ , $p = 0.2$

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

$p (x) = {}_{3}C_{1} \cdot p^{x} \cdot q^{n - x}$

Find the value of ${}_{3}C_{1}$ .

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

${}_{3}C_{1} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Fill in the known values.

$\frac{(3)!}{(1)! (3 - 1)!}$

Simplify.

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Subtract $1$ from $3$ .

$\frac{(3)!}{(1)! 2!}$

Rewrite $(3)!$ as $3 \cdot 2!$ .

$\frac{3 \cdot 2!}{(1)! 2!}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{3 \cdot 2!}{(1)! 2!}$

Rewrite the expression.

$\frac{3}{(1)!}$

Expand $(1)!$ to $1$ .

$\frac{3}{1}$

Divide $3$ by $1$ .

$3$

Fill the known values into the equation.

$3 \cdot (0.2) \cdot {(1 - 0.2)}^{3 - 1}$

Simplify the result.

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Remove parentheses.

$3 \cdot {0.2}^{1} \cdot {(1 - 0.2)}^{3 - 1}$

Evaluate the exponent.

$3 \cdot 0.2 \cdot {(1 - 0.2)}^{3 - 1}$

Multiply $3$ by $0.2$ .

$0.6 \cdot {(1 - 0.2)}^{3 - 1}$

Subtract $0.2$ from $1$ .

$0.6 \cdot {(0.8)}^{3 - 1}$

Subtract $1$ from $3$ .

$0.6 \cdot {(0.8)}^{2}$

Remove parentheses.

$0.6 \cdot {0.8}^{2}$

Raise $0.8$ to the power of $2$ .

$0.6 \cdot 0.64$

Multiply $0.6$ by $0.64$ .

$0.384$

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