Finite Math Examples

$x = 4$ , $n = 4$ , $p = 0.8$

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

$p (x) = {}_{4}C_{4} \cdot p^{x} \cdot q^{n - x}$

Find the value of ${}_{4}C_{4}$ .

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

${}_{4}C_{4} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Fill in the known values.

$\frac{(4)!}{(4)! (4 - 4)!}$

Simplify.

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Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{(4)!}{(4)! (4 - 4)!}$

Rewrite the expression.

$\frac{1}{(4 - 4)!}$

Simplify the denominator.

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Subtract $4$ from $4$ .

$\frac{1}{0!}$

Expand $0!$ to $1$ .

$\frac{1}{1}$

Divide $1$ by $1$ .

$1$

Fill the known values into the equation.

$1 \cdot {(0.8)}^{4} \cdot {(1 - 0.8)}^{4 - 4}$

Simplify the result.

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Multiply ${(0.8)}^{4}$ by $1$ .

${(0.8)}^{4} \cdot {(1 - 0.8)}^{4 - 4}$

Remove parentheses around $0.8$ .

${0.8}^{4} \cdot {(1 - 0.8)}^{4 - 4}$

Raise $0.8$ to the power of $4$ .

$0.4096 \cdot {(1 - 0.8)}^{4 - 4}$

Subtract $0.8$ from $1$ .

$0.4096 \cdot {(0.2)}^{4 - 4}$

Subtract $4$ from $4$ .

$0.4096 \cdot {(0.2)}^{0}$

Remove parentheses around $0.2$ .

$0.4096 \cdot {0.2}^{0}$

Anything raised to $0$ is $1$ .

$0.4096 \cdot 1$

Multiply $0.4096$ by $1$ .

$0.4096$

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