Finite Math Examples

$x < 1$ , $n = 2$ , $p = 0.8$

Step 1

Subtract $0.8$ from $1$ .

$0.2$

Step 2

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, $p (x < 1) = P (x = 0)$ .

$p (x < 1) = P (x = 0)$

Step 3

Find the probability of $p (0)$ .

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

$p (x) = {}_{2}C_{0} \cdot p^{x} \cdot q^{n - x}$

Find the value of ${}_{2}C_{0}$ .

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

${}_{2}C_{0} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Fill in the known values.

$\frac{(2)!}{(0)! (2 - 0)!}$

Simplify.

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Simplify the numerator.

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

$\frac{2 \cdot 1}{(0)! (2 - 0)!}$

Multiply $2$ by $1$ .

$\frac{2}{(0)! (2 - 0)!}$

Simplify the denominator.

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Expand $(0)!$ to $1$ .

$\frac{2}{1 (2 - 0)!}$

Subtract $0$ from $2$ .

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

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

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $1$ .

$\frac{2}{2}$

Divide $2$ by $2$ .

$1$

Fill the known values into the equation.

$1 \cdot {(0.8)}^{0} \cdot {(1 - 0.8)}^{2 - 0}$

Simplify the result.

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Multiply ${(0.8)}^{0}$ by $1$ .

${(0.8)}^{0} \cdot {(1 - 0.8)}^{2 - 0}$

Anything raised to $0$ is $1$ .

$1 \cdot {(1 - 0.8)}^{2 - 0}$

Multiply ${(1 - 0.8)}^{2 - 0}$ by $1$ .

${(1 - 0.8)}^{2 - 0}$

Subtract $0.8$ from $1$ .

${0.2}^{2 - 0}$

Subtract $0$ from $2$ .

${0.2}^{2}$

Raise $0.2$ to the power of $2$ .

$0.04$

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