Finite Math Examples

$x < 3$ , $n = 3$ , $p = 1.21$

Step 1

Subtract $1.21$ from $1$ .

$- 0.21$

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

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

Step 3

Find the probability of $p (0)$ .

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Step 3.1

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

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

Step 3.2

Find the value of ${}_{3}C_{0}$ .

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Step 3.2.1

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

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

Step 3.2.2

Fill in the known values.

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

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 $(3)!$ to $3 \cdot 2 \cdot 1$ .

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

Step 3.2.3.1.2

Multiply $3 \cdot 2 \cdot 1$ .

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Step 3.2.3.1.2.1

Multiply $3$ by $2$ .

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

Step 3.2.3.1.2.2

Multiply $6$ by $1$ .

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

Step 3.2.3.2

Simplify the denominator.

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Step 3.2.3.2.1

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

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

Step 3.2.3.2.2

Subtract $0$ from $3$ .

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

Step 3.2.3.2.3

Expand $(3)!$ to $3 \cdot 2 \cdot 1$ .

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

Step 3.2.3.2.4

Multiply $3 \cdot 2 \cdot 1$ .

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Step 3.2.3.2.4.1

Multiply $3$ by $2$ .

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

Step 3.2.3.2.4.2

Multiply $6$ by $1$ .

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

Step 3.2.3.2.5

Multiply $6$ by $1$ .

$\frac{6}{6}$

Step 3.2.3.3

Divide $6$ by $6$ .

$1$

Step 3.3

Fill the known values into the equation.

$1 \cdot {(1.21)}^{0} \cdot {(1 - 1.21)}^{3 - 0}$

Step 3.4

Simplify the result.

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Step 3.4.1

Multiply ${(1.21)}^{0}$ by $1$ .

${(1.21)}^{0} \cdot {(1 - 1.21)}^{3 - 0}$

Step 3.4.2

Anything raised to $0$ is $1$ .

$1 \cdot {(1 - 1.21)}^{3 - 0}$

Step 3.4.3

Multiply ${(1 - 1.21)}^{3 - 0}$ by $1$ .

${(1 - 1.21)}^{3 - 0}$

Step 3.4.4

Subtract $1.21$ from $1$ .

${(- 0.21)}^{3 - 0}$

Step 3.4.5

Subtract $0$ from $3$ .

${(- 0.21)}^{3}$

Step 3.4.6

Raise $- 0.21$ to the power of $3$ .

$- 0.009261$

Step 4

Find the probability of $p (1)$ .

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Step 4.1

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

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

Step 4.2

Find the value of ${}_{3}C_{1}$ .

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Step 4.2.1

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

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

Step 4.2.2

Fill in the known values.

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

Subtract $1$ from $3$ .

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

Step 4.2.3.2

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

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

Step 4.2.3.3

Cancel the common factor of $2!$ .

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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.

$3 \cdot (1.21) \cdot {(1 - 1.21)}^{3 - 1}$

Step 4.4

Simplify the result.

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Step 4.4.1

Evaluate the exponent.

$3 \cdot 1.21 \cdot {(1 - 1.21)}^{3 - 1}$

Step 4.4.2

Multiply $3$ by $1.21$ .

$3.63 \cdot {(1 - 1.21)}^{3 - 1}$

Step 4.4.3

Subtract $1.21$ from $1$ .

$3.63 \cdot {(- 0.21)}^{3 - 1}$

Step 4.4.4

Subtract $1$ from $3$ .

$3.63 \cdot {(- 0.21)}^{2}$

Step 4.4.5

Raise $- 0.21$ to the power of $2$ .

$3.63 \cdot 0.0441$

Step 4.4.6

Multiply $3.63$ by $0.0441$ .

$0.160083$

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.

$3 \cdot {(1.21)}^{2} \cdot {(1 - 1.21)}^{3 - 2}$

Step 5.4

Simplify the result.

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Step 5.4.1

Raise $1.21$ to the power of $2$ .

$3 \cdot 1.4641 \cdot {(1 - 1.21)}^{3 - 2}$

Step 5.4.2

Multiply $3$ by $1.4641$ .

$4.3923 \cdot {(1 - 1.21)}^{3 - 2}$

Step 5.4.3

Subtract $1.21$ from $1$ .

$4.3923 \cdot {(- 0.21)}^{3 - 2}$

Step 5.4.4

Subtract $2$ from $3$ .

$4.3923 \cdot {(- 0.21)}^{1}$

Step 5.4.5

Evaluate the exponent.

$4.3923 \cdot - 0.21$

Step 5.4.6

Multiply $4.3923$ by $- 0.21$ .

$- 0.922383$

Step 6

The probability $P (x < 3)$ is the sum of the probabilities of all possible $x$ values between $0$ and $n$ . $P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2) = - 0.009261 + 0.160083 - 0.922383 = - 0.771561$ .

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Step 6.1

Add $- 0.009261$ and $0.160083$ .

$p (x < 3) = 0.150822 - 0.922383$

Step 6.2

Subtract $0.922383$ from $0.150822$ .

$p (x < 3) = - 0.771561$