Finite Math Examples

$x > 2$ , $n = 3$ , $p = 0.9$

Subtract $0.9$ from $1$ .

$0.1$

When the value of the number of successes $x$ is given as an interval, then the probability of $x$ is the sum of the probabilities of all possible $x$ values between $0$ and $n$ . In this case, $p (x > 2) = P (x = 3)$ .

$p (x > 2) = P (x = 3)$

Find the probability of $P (3)$ .

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

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

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

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

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

Fill in the known values.

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

Simplify.

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Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify the denominator.

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Subtract $3$ from $3$ .

$\frac{1}{0!}$

Expand $0!$ to $1$ .

$\frac{1}{1}$

Divide $1$ by $1$ .

$1$

Fill the known values into the equation.

$1 \cdot {(0.9)}^{3} \cdot {(1 - 0.9)}^{3 - 3}$

Simplify the result.

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Multiply ${(0.9)}^{3}$ by $1$ .

${(0.9)}^{3} \cdot {(1 - 0.9)}^{3 - 3}$

Remove parentheses around $0.9$ .

${0.9}^{3} \cdot {(1 - 0.9)}^{3 - 3}$

Raise $0.9$ to the power of $3$ .

$0.729 \cdot {(1 - 0.9)}^{3 - 3}$

Subtract $0.9$ from $1$ .

$0.729 \cdot {(0.1)}^{3 - 3}$

Subtract $3$ from $3$ .

$0.729 \cdot {(0.1)}^{0}$

Remove parentheses around $0.1$ .

$0.729 \cdot {0.1}^{0}$

Anything raised to $0$ is $1$ .

$0.729 \cdot 1$

Multiply $0.729$ by $1$ .

$0.729$

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