Finite Math Examples

$x > 1$ , $n = 3$ , $p = 0.4$

Step 1

Subtract $0.4$ from $1$ .

$0.6$

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, $p (x > 1) = P (x = 2) + P (x = 3)$ .

$p (x > 1) = P (x = 2) + P (x = 3)$

Step 3

Find the probability of $P (2)$ .

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Step 3.1

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

$p (x) = {}_{3}C_{2} \cdot p^{x} \cdot q^{n - x}$

Step 3.2

Find the value of ${}_{3}C_{2}$ .

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Step 3.2.1

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

${}_{3}C_{2} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 3.2.2

Fill in the known values.

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

Step 3.2.3

Simplify.

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Step 3.2.3.1

Subtract $2$ from $3$ .

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

Step 3.2.3.2

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

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

Step 3.2.3.3

Cancel the common factor of $2!$ .

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Step 3.2.3.3.1

Cancel the common factor.

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

Step 3.2.3.3.2

Rewrite the expression.

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

Step 3.2.3.4

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

$\frac{3}{1}$

Step 3.2.3.5

Divide $3$ by $1$ .

$3$

Step 3.3

Fill the known values into the equation.

$3 \cdot {(0.4)}^{2} \cdot {(1 - 0.4)}^{3 - 2}$

Step 3.4

Simplify the result.

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Step 3.4.1

Raise $0.4$ to the power of $2$ .

$3 \cdot 0.16 \cdot {(1 - 0.4)}^{3 - 2}$

Step 3.4.2

Multiply $3$ by $0.16$ .

$0.48 \cdot {(1 - 0.4)}^{3 - 2}$

Step 3.4.3

Subtract $0.4$ from $1$ .

$0.48 \cdot {0.6}^{3 - 2}$

Step 3.4.4

Subtract $2$ from $3$ .

$0.48 \cdot {0.6}^{1}$

Step 3.4.5

Evaluate the exponent.

$0.48 \cdot 0.6$

Step 3.4.6

Multiply $0.48$ by $0.6$ .

$0.288$

Step 4

Find the probability of $P (3)$ .

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Step 4.1

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

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Fill in the known values.

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

Cancel the common factor of $(3)!$ .

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Step 4.2.3.1.1

Cancel the common factor.

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

Step 4.2.3.1.2

Rewrite the expression.

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

Step 4.2.3.2

Simplify the denominator.

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Step 4.2.3.2.1

Subtract $3$ from $3$ .

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

Step 4.2.3.2.2

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

$\frac{1}{1}$

Step 4.2.3.3

Divide $1$ by $1$ .

$1$

Step 4.3

Fill the known values into the equation.

$1 \cdot {(0.4)}^{3} \cdot {(1 - 0.4)}^{3 - 3}$

Step 4.4

Simplify the result.

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Step 4.4.1

Multiply ${(0.4)}^{3}$ by $1$ .

${(0.4)}^{3} \cdot {(1 - 0.4)}^{3 - 3}$

Step 4.4.2

Raise $0.4$ to the power of $3$ .

$0.064 \cdot {(1 - 0.4)}^{3 - 3}$

Step 4.4.3

Subtract $0.4$ from $1$ .

$0.064 \cdot {0.6}^{3 - 3}$

Step 4.4.4

Subtract $3$ from $3$ .

$0.064 \cdot {0.6}^{0}$

Step 4.4.5

Anything raised to $0$ is $1$ .

$0.064 \cdot 1$

Step 4.4.6

Multiply $0.064$ by $1$ .

$0.064$

Step 5

Add $0.288$ and $0.064$ .

$p (x > 1) = 0.352$

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