Finite Math Examples

x < 3

$x < 3$ ,

n = 3

$n = 3$ ,

p = 0.4

$p = 0.4$

Step 1

Subtract

0.4

$0.4$ from

1

$1$ .

0.6

$0.6$

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 < 3) = P (x = 0) + P (x = 1) + P (x = 2)

$p (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)$ .

p (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)

$p (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)$

Step 3

Find the probability of

p (0)

$p (0)$ .

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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 0 \cdot p^{x} \cdot q^{n - x}

$p (x) = {}_{3}C_{0} \cdot p^{x} \cdot q^{n - x}$

Step 3.2

Find the value of

3 C 0

${}_{3}C_{0}$ .

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

${}_{3}C_{0} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 3.2.2

Fill in the known values.

\frac{(3)!}{(0)! (3 - 0)!}

$\frac{(3)!}{(0)! (3 - 0)!}$

Step 3.2.3

Simplify.

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Step 3.2.3.1

Simplify the numerator.

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Step 3.2.3.1.1

Expand

(3)!

$(3)!$ to

3 \cdot 2 \cdot 1

$3 \cdot 2 \cdot 1$ .

\frac{3 \cdot 2 \cdot 1}{(0)! (3 - 0)!}

$\frac{3 \cdot 2 \cdot 1}{(0)! (3 - 0)!}$

Step 3.2.3.1.2

Multiply

3 \cdot 2 \cdot 1

$3 \cdot 2 \cdot 1$ .

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Step 3.2.3.1.2.1

Multiply

3

$3$ by

2

$2$ .

\frac{6 \cdot 1}{(0)! (3 - 0)!}

$\frac{6 \cdot 1}{(0)! (3 - 0)!}$

Step 3.2.3.1.2.2

Multiply

6

$6$ by

1

$1$ .

\frac{6}{(0)! (3 - 0)!}

$\frac{6}{(0)! (3 - 0)!}$

\frac{6}{(0)! (3 - 0)!}

$\frac{6}{(0)! (3 - 0)!}$

\frac{6}{(0)! (3 - 0)!}

$\frac{6}{(0)! (3 - 0)!}$

Step 3.2.3.2

Simplify the denominator.

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Step 3.2.3.2.1

Expand

(0)!

$(0)!$ to

1

$1$ .

\frac{6}{1 (3 - 0)!}

$\frac{6}{1 (3 - 0)!}$

Step 3.2.3.2.2

Subtract

0

$0$ from

3

$3$ .

\frac{6}{1 (3)!}

$\frac{6}{1 (3)!}$

Step 3.2.3.2.3

Expand

(3)!

$(3)!$ to

3 \cdot 2 \cdot 1

$3 \cdot 2 \cdot 1$ .

\frac{6}{1 (3 \cdot 2 \cdot 1)}

$\frac{6}{1 (3 \cdot 2 \cdot 1)}$

Step 3.2.3.2.4

Multiply

3 \cdot 2 \cdot 1

$3 \cdot 2 \cdot 1$ .

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Step 3.2.3.2.4.1

Multiply

3

$3$ by

2

$2$ .

\frac{6}{1 (6 \cdot 1)}

$\frac{6}{1 (6 \cdot 1)}$

Step 3.2.3.2.4.2

Multiply

6

$6$ by

1

$1$ .

\frac{6}{1 \cdot 6}

$\frac{6}{1 \cdot 6}$

\frac{6}{1 \cdot 6}

$\frac{6}{1 \cdot 6}$

Step 3.2.3.2.5

Multiply

6

$6$ by

1

$1$ .

\frac{6}{6}

$\frac{6}{6}$

\frac{6}{6}

$\frac{6}{6}$

Step 3.2.3.3

Divide

6

$6$ by

6

$6$ .

1

$1$

1

$1$

1

$1$

Step 3.3

Fill the known values into the equation.

1 \cdot {(0.4)}^{0} \cdot {(1 - 0.4)}^{3 - 0}

$1 \cdot {(0.4)}^{0} \cdot {(1 - 0.4)}^{3 - 0}$

Step 3.4

Simplify the result.

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Step 3.4.1

Multiply

{(0.4)}^{0}

${(0.4)}^{0}$ by

1

$1$ .

{(0.4)}^{0} \cdot {(1 - 0.4)}^{3 - 0}

${(0.4)}^{0} \cdot {(1 - 0.4)}^{3 - 0}$

Step 3.4.2

Anything raised to

0

$0$ is

1

$1$ .

1 \cdot {(1 - 0.4)}^{3 - 0}

$1 \cdot {(1 - 0.4)}^{3 - 0}$

Step 3.4.3

Multiply

{(1 - 0.4)}^{3 - 0}

${(1 - 0.4)}^{3 - 0}$ by

1

$1$ .

{(1 - 0.4)}^{3 - 0}

${(1 - 0.4)}^{3 - 0}$

Step 3.4.4

Subtract

0.4

$0.4$ from

1

$1$ .

{0.6}^{3 - 0}

${0.6}^{3 - 0}$

Step 3.4.5

Subtract

0

$0$ from

3

$3$ .

{0.6}^{3}

${0.6}^{3}$

Step 3.4.6

Raise

0.6

$0.6$ to the power of

3

$3$ .

0.216

$0.216$

0.216

$0.216$

0.216

$0.216$

Step 4

Find the probability of

p (1)

$p (1)$ .

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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 1 \cdot p^{x} \cdot q^{n - x}

$p (x) = {}_{3}C_{1} \cdot p^{x} \cdot q^{n - x}$

Step 4.2

Find the value of

3 C 1

${}_{3}C_{1}$ .

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Step 4.2.1

Find the number of possible unordered combinations when

r

$r$ items are selected from

n

$n$ available items.

3 C 1 = n C r = \frac{n!}{(r)! (n - r)!}

${}_{3}C_{1} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 4.2.2

Fill in the known values.

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

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

Subtract

1

$1$ from

3

$3$ .

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

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

Step 4.2.3.2

Rewrite

(3)!

$(3)!$ as

3 \cdot 2!

$3 \cdot 2!$ .

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

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

Step 4.2.3.3

Cancel the common factor of

2!

$2!$ .

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Step 4.2.3.3.1

Cancel the common factor.

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

Step 4.2.3.3.2

Rewrite the expression.

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

Step 4.2.3.4

Expand

$(1)!$ to

$1$ .

$\frac{3}{1}$

Step 4.2.3.5

Divide

$3$ by

$1$ .

$3$

Step 4.3

Fill the known values into the equation.

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

Step 4.4

Simplify the result.

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Step 4.4.1

Evaluate the exponent.

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

Step 4.4.2

Multiply

$3$ by

$0.4$ .

$1.2 \cdot {(1 - 0.4)}^{3 - 1}$

Step 4.4.3

Subtract

$0.4$ from

$1$ .

$1.2 \cdot {0.6}^{3 - 1}$

Step 4.4.4

Subtract

$1$ from

$3$ .

$1.2 \cdot {0.6}^{2}$

Step 4.4.5

Raise

$0.6$ to the power of

$2$ .

$1.2 \cdot 0.36$

Step 4.4.6

Multiply

$1.2$ by

$0.36$ .

$0.432$

Step 5

Find the probability of

$p (2)$ .

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Step 5.1

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 5.2

Find the value of

${}_{3}C_{2}$ .

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Step 5.2.1

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 5.2.2

Fill in the known values.

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

Step 5.2.3

Simplify.

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Step 5.2.3.1

Subtract

$2$ from

$3$ .

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

Step 5.2.3.2

Rewrite

$(3)!$ as

$3 \cdot 2!$ .

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

Step 5.2.3.3

Cancel the common factor of

$2!$ .

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Step 5.2.3.3.1

Cancel the common factor.

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

Step 5.2.3.3.2

Rewrite the expression.

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

Step 5.2.3.4

Expand

$(1)!$ to

$1$ .

$\frac{3}{1}$

Step 5.2.3.5

Divide

$3$ by

$1$ .

$3$

Step 5.3

Fill the known values into the equation.

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

Step 5.4

Simplify the result.

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Step 5.4.1

Raise

$0.4$ to the power of

$2$ .

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

Step 5.4.2

Multiply

$3$ by

$0.16$ .

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

Step 5.4.3

Subtract

$0.4$ from

$1$ .

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

Step 5.4.4

Subtract

$2$ from

$3$ .

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

Step 5.4.5

Evaluate the exponent.

$0.48 \cdot 0.6$

Step 5.4.6

Multiply

$0.48$ by

$0.6$ .

$0.288$

Step 6

The probability

$P (x < 3)$ is the sum of the probabilities of all possible

$x$ values between

$0$ and

$n$ .

$P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2) = 0.216 + 0.432 + 0.288 = 0.936$ .

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Step 6.1

Add

$0.216$ and

$0.432$ .

$p (x < 3) = 0.648 + 0.288$

Step 6.2

Add

$0.648$ and

$0.288$ .

$p (x < 3) = 0.936$

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