Finite Math Examples

x > 2

$x > 2$ ,

n = 3

$n = 3$ ,

p = 0.9

$p = 0.9$

Step 1

Subtract

0.9

$0.9$ from

1

$1$ .

0.1

$0.1$

Step 2

When the value of the number of successes

x

$x$ is given as an interval, then the probability of

x

$x$ is the sum of the probabilities of all possible

x

$x$ values between

0

$0$ and

n

$n$ . In this case,

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

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

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

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

Step 3

Find the probability of

P (3)

$P (3)$ .

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Step 3.1

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}

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

Step 3.2

Find the value of

3 C 3

${}_{3}C_{3}$ .

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Step 3.2.1

Find the number of possible unordered combinations when

r

$r$ items are selected from

n

$n$ available items.

3 C 3 = n C r = \frac{n!}{(r)! (n - r)!}

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

Step 3.2.2

Fill in the known values.

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

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

Step 3.2.3

Simplify.

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Step 3.2.3.1

Cancel the common factor of

(3)!

$(3)!$ .

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Step 3.2.3.1.1

Cancel the common factor.

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

Step 3.2.3.1.2

Rewrite the expression.

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

Step 3.2.3.2

Simplify the denominator.

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Step 3.2.3.2.1

Subtract

$3$ from

$3$ .

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

Step 3.2.3.2.2

Expand

$(0)!$ to

$1$ .

$\frac{1}{1}$

Step 3.2.3.3

Divide

$1$ by

$1$ .

$1$

Step 3.3

Fill the known values into the equation.

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

Step 3.4

Simplify the result.

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Step 3.4.1

Multiply

${(0.9)}^{3}$ by

$1$ .

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

Step 3.4.2

Raise

$0.9$ to the power of

$3$ .

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

Step 3.4.3

Subtract

$0.9$ from

$1$ .

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

Step 3.4.4

Subtract

$3$ from

$3$ .

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

Step 3.4.5

Anything raised to

$0$ is

$1$ .

$0.729 \cdot 1$

Step 3.4.6

Multiply

$0.729$ by

$1$ .

$0.729$

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