Finite Math Examples

$x > 3$ , $n = 5$ , $p = 0.5$

Step 1

Subtract $0.5$ from $1$ .

$0.5$

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, $p (x > 3) = P (x = 4) + P (x = 5)$ .

$p (x > 3) = P (x = 4) + P (x = 5)$

Step 3

Find the probability of $P (4)$ .

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Step 3.1

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

$p (x) = {}_{5}C_{4} \cdot p^{x} \cdot q^{n - x}$

Step 3.2

Find the value of ${}_{5}C_{4}$ .

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Step 3.2.1

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

${}_{5}C_{4} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 3.2.2

Fill in the known values.

$\frac{(5)!}{(4)! (5 - 4)!}$

Step 3.2.3

Simplify.

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Step 3.2.3.1

Subtract $4$ from $5$ .

$\frac{(5)!}{(4)! (1)!}$

Step 3.2.3.2

Rewrite $(5)!$ as $5 \cdot 4!$ .

$\frac{5 \cdot 4!}{(4)! (1)!}$

Step 3.2.3.3

Cancel the common factor of $4!$ .

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Step 3.2.3.3.1

Cancel the common factor.

$\frac{5 \cdot 4!}{(4)! (1)!}$

Step 3.2.3.3.2

Rewrite the expression.

$\frac{5}{(1)!}$

Step 3.2.3.4

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

$\frac{5}{1}$

Step 3.2.3.5

Divide $5$ by $1$ .

$5$

Step 3.3

Fill the known values into the equation.

$5 \cdot {(0.5)}^{4} \cdot {(1 - 0.5)}^{5 - 4}$

Step 3.4

Simplify the result.

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Step 3.4.1

Raise $0.5$ to the power of $4$ .

$5 \cdot 0.0625 \cdot {(1 - 0.5)}^{5 - 4}$

Step 3.4.2

Multiply $5$ by $0.0625$ .

$0.3125 \cdot {(1 - 0.5)}^{5 - 4}$

Step 3.4.3

Subtract $0.5$ from $1$ .

$0.3125 \cdot {0.5}^{5 - 4}$

Step 3.4.4

Subtract $4$ from $5$ .

$0.3125 \cdot {0.5}^{1}$

Step 3.4.5

Evaluate the exponent.

$0.3125 \cdot 0.5$

Step 3.4.6

Multiply $0.3125$ by $0.5$ .

$0.15625$

Step 4

Find the probability of $P (5)$ .

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Step 4.1

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

$p (x) = {}_{5}C_{5} \cdot p^{x} \cdot q^{n - x}$

Step 4.2

Find the value of ${}_{5}C_{5}$ .

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Step 4.2.1

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

${}_{5}C_{5} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 4.2.2

Fill in the known values.

$\frac{(5)!}{(5)! (5 - 5)!}$

Step 4.2.3

Simplify.

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Step 4.2.3.1

Cancel the common factor of $(5)!$ .

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Step 4.2.3.1.1

Cancel the common factor.

$\frac{(5)!}{(5)! (5 - 5)!}$

Step 4.2.3.1.2

Rewrite the expression.

$\frac{1}{(5 - 5)!}$

Step 4.2.3.2

Simplify the denominator.

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Step 4.2.3.2.1

Subtract $5$ from $5$ .

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

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.

$1 \cdot {(0.5)}^{5} \cdot {(1 - 0.5)}^{5 - 5}$

Step 4.4

Simplify the result.

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Step 4.4.1

Multiply ${(0.5)}^{5}$ by $1$ .

${(0.5)}^{5} \cdot {(1 - 0.5)}^{5 - 5}$

Step 4.4.2

Raise $0.5$ to the power of $5$ .

$0.03125 \cdot {(1 - 0.5)}^{5 - 5}$

Step 4.4.3

Subtract $0.5$ from $1$ .

$0.03125 \cdot {0.5}^{5 - 5}$

Step 4.4.4

Subtract $5$ from $5$ .

$0.03125 \cdot {0.5}^{0}$

Step 4.4.5

Anything raised to $0$ is $1$ .

$0.03125 \cdot 1$

Step 4.4.6

Multiply $0.03125$ by $1$ .

$0.03125$

Step 5

Add $0.15625$ and $0.03125$ .

$p (x > 3) = 0.1875$