Finite Math Examples

$x > 0$ , $n = 12$ , $p = 0.3$

Step 1

Subtract $0.3$ from $1$ .

$0.7$

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, $p (x > 0) = P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6) + P (x = 7) + P (x = 8) + P (x = 9) + P (x = 10) + P (x = 11) + P (x = 12)$ .

$p (x > 0) = P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6) + P (x = 7) + P (x = 8) + P (x = 9) + P (x = 10) + P (x = 11) + P (x = 12)$

Step 3

Find the probability of $P (1)$ .

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Step 3.1

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

$p (x) = {}_{12}C_{1} \cdot p^{x} \cdot q^{n - x}$

Step 3.2

Find the value of ${}_{12}C_{1}$ .

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Step 3.2.1

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

${}_{12}C_{1} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 3.2.2

Fill in the known values.

$\frac{(12)!}{(1)! (12 - 1)!}$

Step 3.2.3

Simplify.

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Step 3.2.3.1

Subtract $1$ from $12$ .

$\frac{(12)!}{(1)! (11)!}$

Step 3.2.3.2

Rewrite $(12)!$ as $12 \cdot 11!$ .

$\frac{12 \cdot 11!}{(1)! (11)!}$

Step 3.2.3.3

Cancel the common factor of $11!$ .

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Step 3.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11!}{(1)! (11)!}$

Step 3.2.3.3.2

Rewrite the expression.

$\frac{12}{(1)!}$

Step 3.2.3.4

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

$\frac{12}{1}$

Step 3.2.3.5

Divide $12$ by $1$ .

$12$

Step 3.3

Fill the known values into the equation.

$12 \cdot (0.3) \cdot {(1 - 0.3)}^{12 - 1}$

Step 3.4

Simplify the result.

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Step 3.4.1

Evaluate the exponent.

$12 \cdot 0.3 \cdot {(1 - 0.3)}^{12 - 1}$

Step 3.4.2

Multiply $12$ by $0.3$ .

$3.6 \cdot {(1 - 0.3)}^{12 - 1}$

Step 3.4.3

Subtract $0.3$ from $1$ .

$3.6 \cdot {0.7}^{12 - 1}$

Step 3.4.4

Subtract $1$ from $12$ .

$3.6 \cdot {0.7}^{11}$

Step 3.4.5

Raise $0.7$ to the power of $11$ .

$3.6 \cdot 0.01977326$

Step 3.4.6

Multiply $3.6$ by $0.01977326$ .

$0.07118376$

Step 4

Find the probability of $P (2)$ .

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Step 4.1

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

$p (x) = {}_{12}C_{2} \cdot p^{x} \cdot q^{n - x}$

Step 4.2

Find the value of ${}_{12}C_{2}$ .

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Step 4.2.1

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

${}_{12}C_{2} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 4.2.2

Fill in the known values.

$\frac{(12)!}{(2)! (12 - 2)!}$

Step 4.2.3

Simplify.

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Step 4.2.3.1

Subtract $2$ from $12$ .

$\frac{(12)!}{(2)! (10)!}$

Step 4.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10!$ .

$\frac{12 \cdot 11 \cdot 10!}{(2)! (10)!}$

Step 4.2.3.3

Reduce the expression by cancelling the common factors.

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Step 4.2.3.3.1

Cancel the common factor of $10!$ .

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Step 4.2.3.3.1.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10!}{(2)! (10)!}$

Step 4.2.3.3.1.2

Rewrite the expression.

$\frac{12 \cdot 11}{(2)!}$

Step 4.2.3.3.2

Multiply $12$ by $11$ .

$\frac{132}{(2)!}$

Step 4.2.3.4

Simplify the denominator.

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Step 4.2.3.4.1

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

$\frac{132}{2 \cdot 1}$

Step 4.2.3.4.2

Multiply $2$ by $1$ .

$\frac{132}{2}$

Step 4.2.3.5

Divide $132$ by $2$ .

$66$

Step 4.3

Fill the known values into the equation.

$66 \cdot {(0.3)}^{2} \cdot {(1 - 0.3)}^{12 - 2}$

Step 4.4

Simplify the result.

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Step 4.4.1

Raise $0.3$ to the power of $2$ .

$66 \cdot 0.09 \cdot {(1 - 0.3)}^{12 - 2}$

Step 4.4.2

Multiply $66$ by $0.09$ .

$5.94 \cdot {(1 - 0.3)}^{12 - 2}$

Step 4.4.3

Subtract $0.3$ from $1$ .

$5.94 \cdot {0.7}^{12 - 2}$

Step 4.4.4

Subtract $2$ from $12$ .

$5.94 \cdot {0.7}^{10}$

Step 4.4.5

Raise $0.7$ to the power of $10$ .

$5.94 \cdot 0.02824752$

Step 4.4.6

Multiply $5.94$ by $0.02824752$ .

$0.16779029$

Step 5

Find the probability of $P (3)$ .

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Step 5.1

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

$p (x) = {}_{12}C_{3} \cdot p^{x} \cdot q^{n - x}$

Step 5.2

Find the value of ${}_{12}C_{3}$ .

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Step 5.2.1

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

${}_{12}C_{3} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 5.2.2

Fill in the known values.

$\frac{(12)!}{(3)! (12 - 3)!}$

Step 5.2.3

Simplify.

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Step 5.2.3.1

Subtract $3$ from $12$ .

$\frac{(12)!}{(3)! (9)!}$

Step 5.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9!}{(3)! (9)!}$

Step 5.2.3.3

Cancel the common factor of $9!$ .

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Step 5.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9!}{(3)! (9)!}$

Step 5.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10}{(3)!}$

Step 5.2.3.4

Simplify the numerator.

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Step 5.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10}{(3)!}$

Step 5.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320}{(3)!}$

Step 5.2.3.5

Simplify the denominator.

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Step 5.2.3.5.1

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

$\frac{1320}{3 \cdot 2 \cdot 1}$

Step 5.2.3.5.2

Multiply $3 \cdot 2 \cdot 1$ .

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Step 5.2.3.5.2.1

Multiply $3$ by $2$ .

$\frac{1320}{6 \cdot 1}$

Step 5.2.3.5.2.2

Multiply $6$ by $1$ .

$\frac{1320}{6}$

Step 5.2.3.6

Divide $1320$ by $6$ .

$220$

Step 5.3

Fill the known values into the equation.

$220 \cdot {(0.3)}^{3} \cdot {(1 - 0.3)}^{12 - 3}$

Step 5.4

Simplify the result.

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Step 5.4.1

Raise $0.3$ to the power of $3$ .

$220 \cdot 0.027 \cdot {(1 - 0.3)}^{12 - 3}$

Step 5.4.2

Multiply $220$ by $0.027$ .

$5.94 \cdot {(1 - 0.3)}^{12 - 3}$

Step 5.4.3

Subtract $0.3$ from $1$ .

$5.94 \cdot {0.7}^{12 - 3}$

Step 5.4.4

Subtract $3$ from $12$ .

$5.94 \cdot {0.7}^{9}$

Step 5.4.5

Raise $0.7$ to the power of $9$ .

$5.94 \cdot 0.0403536$

Step 5.4.6

Multiply $5.94$ by $0.0403536$ .

$0.23970042$

Step 6

Find the probability of $P (4)$ .

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Step 6.1

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

$p (x) = {}_{12}C_{4} \cdot p^{x} \cdot q^{n - x}$

Step 6.2

Find the value of ${}_{12}C_{4}$ .

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Step 6.2.1

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

${}_{12}C_{4} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 6.2.2

Fill in the known values.

$\frac{(12)!}{(4)! (12 - 4)!}$

Step 6.2.3

Simplify.

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Step 6.2.3.1

Subtract $4$ from $12$ .

$\frac{(12)!}{(4)! (8)!}$

Step 6.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{(4)! (8)!}$

Step 6.2.3.3

Cancel the common factor of $8!$ .

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Step 6.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{(4)! (8)!}$

Step 6.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10 \cdot 9}{(4)!}$

Step 6.2.3.4

Simplify the numerator.

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Step 6.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10 \cdot 9}{(4)!}$

Step 6.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320 \cdot 9}{(4)!}$

Step 6.2.3.4.3

Multiply $1320$ by $9$ .

$\frac{11880}{(4)!}$

Step 6.2.3.5

Simplify the denominator.

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Step 6.2.3.5.1

Expand $(4)!$ to $4 \cdot 3 \cdot 2 \cdot 1$ .

$\frac{11880}{4 \cdot 3 \cdot 2 \cdot 1}$

Step 6.2.3.5.2

Multiply $4 \cdot 3 \cdot 2 \cdot 1$ .

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Step 6.2.3.5.2.1

Multiply $4$ by $3$ .

$\frac{11880}{12 \cdot 2 \cdot 1}$

Step 6.2.3.5.2.2

Multiply $12$ by $2$ .

$\frac{11880}{24 \cdot 1}$

Step 6.2.3.5.2.3

Multiply $24$ by $1$ .

$\frac{11880}{24}$

Step 6.2.3.6

Divide $11880$ by $24$ .

$495$

Step 6.3

Fill the known values into the equation.

$495 \cdot {(0.3)}^{4} \cdot {(1 - 0.3)}^{12 - 4}$

Step 6.4

Simplify the result.

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Step 6.4.1

Raise $0.3$ to the power of $4$ .

$495 \cdot 0.0081 \cdot {(1 - 0.3)}^{12 - 4}$

Step 6.4.2

Multiply $495$ by $0.0081$ .

$4.0095 \cdot {(1 - 0.3)}^{12 - 4}$

Step 6.4.3

Subtract $0.3$ from $1$ .

$4.0095 \cdot {0.7}^{12 - 4}$

Step 6.4.4

Subtract $4$ from $12$ .

$4.0095 \cdot {0.7}^{8}$

Step 6.4.5

Raise $0.7$ to the power of $8$ .

$4.0095 \cdot 0.05764801$

Step 6.4.6

Multiply $4.0095$ by $0.05764801$ .

$0.23113969$

Step 7

Find the probability of $P (5)$ .

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Step 7.1

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

$p (x) = {}_{12}C_{5} \cdot p^{x} \cdot q^{n - x}$

Step 7.2

Find the value of ${}_{12}C_{5}$ .

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Step 7.2.1

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

${}_{12}C_{5} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 7.2.2

Fill in the known values.

$\frac{(12)!}{(5)! (12 - 5)!}$

Step 7.2.3

Simplify.

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Step 7.2.3.1

Subtract $5$ from $12$ .

$\frac{(12)!}{(5)! (7)!}$

Step 7.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}{(5)! (7)!}$

Step 7.2.3.3

Cancel the common factor of $7!$ .

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Step 7.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}{(5)! (7)!}$

Step 7.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{(5)!}$

Step 7.2.3.4

Simplify the numerator.

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Step 7.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10 \cdot 9 \cdot 8}{(5)!}$

Step 7.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320 \cdot 9 \cdot 8}{(5)!}$

Step 7.2.3.4.3

Multiply $1320$ by $9$ .

$\frac{11880 \cdot 8}{(5)!}$

Step 7.2.3.4.4

Multiply $11880$ by $8$ .

$\frac{95040}{(5)!}$

Step 7.2.3.5

Simplify the denominator.

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Step 7.2.3.5.1

Expand $(5)!$ to $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .

$\frac{95040}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Step 7.2.3.5.2

Multiply $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .

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Step 7.2.3.5.2.1

Multiply $5$ by $4$ .

$\frac{95040}{20 \cdot 3 \cdot 2 \cdot 1}$

Step 7.2.3.5.2.2

Multiply $20$ by $3$ .

$\frac{95040}{60 \cdot 2 \cdot 1}$

Step 7.2.3.5.2.3

Multiply $60$ by $2$ .

$\frac{95040}{120 \cdot 1}$

Step 7.2.3.5.2.4

Multiply $120$ by $1$ .

$\frac{95040}{120}$

Step 7.2.3.6

Divide $95040$ by $120$ .

$792$

Step 7.3

Fill the known values into the equation.

$792 \cdot {(0.3)}^{5} \cdot {(1 - 0.3)}^{12 - 5}$

Step 7.4

Simplify the result.

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Step 7.4.1

Raise $0.3$ to the power of $5$ .

$792 \cdot 0.00243 \cdot {(1 - 0.3)}^{12 - 5}$

Step 7.4.2

Multiply $792$ by $0.00243$ .

$1.92456 \cdot {(1 - 0.3)}^{12 - 5}$

Step 7.4.3

Subtract $0.3$ from $1$ .

$1.92456 \cdot {0.7}^{12 - 5}$

Step 7.4.4

Subtract $5$ from $12$ .

$1.92456 \cdot {0.7}^{7}$

Step 7.4.5

Raise $0.7$ to the power of $7$ .

$1.92456 \cdot 0.0823543$

Step 7.4.6

Multiply $1.92456$ by $0.0823543$ .

$0.15849579$

Step 8

Find the probability of $P (6)$ .

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Step 8.1

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

$p (x) = {}_{12}C_{6} \cdot p^{x} \cdot q^{n - x}$

Step 8.2

Find the value of ${}_{12}C_{6}$ .

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Step 8.2.1

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

${}_{12}C_{6} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 8.2.2

Fill in the known values.

$\frac{(12)!}{(6)! (12 - 6)!}$

Step 8.2.3

Simplify.

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Step 8.2.3.1

Subtract $6$ from $12$ .

$\frac{(12)!}{(6)! (6)!}$

Step 8.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{(6)! (6)!}$

Step 8.2.3.3

Cancel the common factor of $6!$ .

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Step 8.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{(6)! (6)!}$

Step 8.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{(6)!}$

Step 8.2.3.4

Simplify the numerator.

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Step 8.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{(6)!}$

Step 8.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320 \cdot 9 \cdot 8 \cdot 7}{(6)!}$

Step 8.2.3.4.3

Multiply $1320$ by $9$ .

$\frac{11880 \cdot 8 \cdot 7}{(6)!}$

Step 8.2.3.4.4

Multiply $11880$ by $8$ .

$\frac{95040 \cdot 7}{(6)!}$

Step 8.2.3.4.5

Multiply $95040$ by $7$ .

$\frac{665280}{(6)!}$

Step 8.2.3.5

Simplify the denominator.

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Step 8.2.3.5.1

Expand $(6)!$ to $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .

$\frac{665280}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Step 8.2.3.5.2

Multiply $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .

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Step 8.2.3.5.2.1

Multiply $6$ by $5$ .

$\frac{665280}{30 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Step 8.2.3.5.2.2

Multiply $30$ by $4$ .

$\frac{665280}{120 \cdot 3 \cdot 2 \cdot 1}$

Step 8.2.3.5.2.3

Multiply $120$ by $3$ .

$\frac{665280}{360 \cdot 2 \cdot 1}$

Step 8.2.3.5.2.4

Multiply $360$ by $2$ .

$\frac{665280}{720 \cdot 1}$

Step 8.2.3.5.2.5

Multiply $720$ by $1$ .

$\frac{665280}{720}$

Step 8.2.3.6

Divide $665280$ by $720$ .

$924$

Step 8.3

Fill the known values into the equation.

$924 \cdot {(0.3)}^{6} \cdot {(1 - 0.3)}^{12 - 6}$

Step 8.4

Simplify the result.

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Step 8.4.1

Raise $0.3$ to the power of $6$ .

$924 \cdot 0.000729 \cdot {(1 - 0.3)}^{12 - 6}$

Step 8.4.2

Multiply $924$ by $0.000729$ .

$0.673596 \cdot {(1 - 0.3)}^{12 - 6}$

Step 8.4.3

Subtract $0.3$ from $1$ .

$0.673596 \cdot {0.7}^{12 - 6}$

Step 8.4.4

Subtract $6$ from $12$ .

$0.673596 \cdot {0.7}^{6}$

Step 8.4.5

Raise $0.7$ to the power of $6$ .

$0.673596 \cdot 0.117649$

Step 8.4.6

Multiply $0.673596$ by $0.117649$ .

$0.07924789$

Step 9

Find the probability of $P (7)$ .

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Step 9.1

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

$p (x) = {}_{12}C_{7} \cdot p^{x} \cdot q^{n - x}$

Step 9.2

Find the value of ${}_{12}C_{7}$ .

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Step 9.2.1

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

${}_{12}C_{7} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 9.2.2

Fill in the known values.

$\frac{(12)!}{(7)! (12 - 7)!}$

Step 9.2.3

Simplify.

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Step 9.2.3.1

Subtract $7$ from $12$ .

$\frac{(12)!}{(7)! (5)!}$

Step 9.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}{(7)! (5)!}$

Step 9.2.3.3

Cancel the common factor of $7!$ .

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Step 9.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}{(7)! (5)!}$

Step 9.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{(5)!}$

Step 9.2.3.4

Simplify the numerator.

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Step 9.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10 \cdot 9 \cdot 8}{(5)!}$

Step 9.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320 \cdot 9 \cdot 8}{(5)!}$

Step 9.2.3.4.3

Multiply $1320$ by $9$ .

$\frac{11880 \cdot 8}{(5)!}$

Step 9.2.3.4.4

Multiply $11880$ by $8$ .

$\frac{95040}{(5)!}$

Step 9.2.3.5

Simplify the denominator.

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Step 9.2.3.5.1

Expand $(5)!$ to $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .

$\frac{95040}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Step 9.2.3.5.2

Multiply $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .

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Step 9.2.3.5.2.1

Multiply $5$ by $4$ .

$\frac{95040}{20 \cdot 3 \cdot 2 \cdot 1}$

Step 9.2.3.5.2.2

Multiply $20$ by $3$ .

$\frac{95040}{60 \cdot 2 \cdot 1}$

Step 9.2.3.5.2.3

Multiply $60$ by $2$ .

$\frac{95040}{120 \cdot 1}$

Step 9.2.3.5.2.4

Multiply $120$ by $1$ .

$\frac{95040}{120}$

Step 9.2.3.6

Divide $95040$ by $120$ .

$792$

Step 9.3

Fill the known values into the equation.

$792 \cdot {(0.3)}^{7} \cdot {(1 - 0.3)}^{12 - 7}$

Step 9.4

Simplify the result.

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Step 9.4.1

Raise $0.3$ to the power of $7$ .

$792 \cdot 0.0002187 \cdot {(1 - 0.3)}^{12 - 7}$

Step 9.4.2

Multiply $792$ by $0.0002187$ .

$0.1732104 \cdot {(1 - 0.3)}^{12 - 7}$

Step 9.4.3

Subtract $0.3$ from $1$ .

$0.1732104 \cdot {0.7}^{12 - 7}$

Step 9.4.4

Subtract $7$ from $12$ .

$0.1732104 \cdot {0.7}^{5}$

Step 9.4.5

Raise $0.7$ to the power of $5$ .

$0.1732104 \cdot 0.16807$

Step 9.4.6

Multiply $0.1732104$ by $0.16807$ .

$0.02911147$

Step 10

Find the probability of $P (8)$ .

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Step 10.1

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

$p (x) = {}_{12}C_{8} \cdot p^{x} \cdot q^{n - x}$

Step 10.2

Find the value of ${}_{12}C_{8}$ .

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Step 10.2.1

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

${}_{12}C_{8} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 10.2.2

Fill in the known values.

$\frac{(12)!}{(8)! (12 - 8)!}$

Step 10.2.3

Simplify.

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Step 10.2.3.1

Subtract $8$ from $12$ .

$\frac{(12)!}{(8)! (4)!}$

Step 10.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{(8)! (4)!}$

Step 10.2.3.3

Cancel the common factor of $8!$ .

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Step 10.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{(8)! (4)!}$

Step 10.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10 \cdot 9}{(4)!}$

Step 10.2.3.4

Simplify the numerator.

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Step 10.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10 \cdot 9}{(4)!}$

Step 10.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320 \cdot 9}{(4)!}$

Step 10.2.3.4.3

Multiply $1320$ by $9$ .

$\frac{11880}{(4)!}$

Step 10.2.3.5

Simplify the denominator.

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Step 10.2.3.5.1

Expand $(4)!$ to $4 \cdot 3 \cdot 2 \cdot 1$ .

$\frac{11880}{4 \cdot 3 \cdot 2 \cdot 1}$

Step 10.2.3.5.2

Multiply $4 \cdot 3 \cdot 2 \cdot 1$ .

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Step 10.2.3.5.2.1

Multiply $4$ by $3$ .

$\frac{11880}{12 \cdot 2 \cdot 1}$

Step 10.2.3.5.2.2

Multiply $12$ by $2$ .

$\frac{11880}{24 \cdot 1}$

Step 10.2.3.5.2.3

Multiply $24$ by $1$ .

$\frac{11880}{24}$

Step 10.2.3.6

Divide $11880$ by $24$ .

$495$

Step 10.3

Fill the known values into the equation.

$495 \cdot {(0.3)}^{8} \cdot {(1 - 0.3)}^{12 - 8}$

Step 10.4

Simplify the result.

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Step 10.4.1

Raise $0.3$ to the power of $8$ .

$495 \cdot 0.00006561 \cdot {(1 - 0.3)}^{12 - 8}$

Step 10.4.2

Multiply $495$ by $0.00006561$ .

$0.03247695 \cdot {(1 - 0.3)}^{12 - 8}$

Step 10.4.3

Subtract $0.3$ from $1$ .

$0.03247695 \cdot {0.7}^{12 - 8}$

Step 10.4.4

Subtract $8$ from $12$ .

$0.03247695 \cdot {0.7}^{4}$

Step 10.4.5

Raise $0.7$ to the power of $4$ .

$0.03247695 \cdot 0.2401$

Step 10.4.6

Multiply $0.03247695$ by $0.2401$ .

$0.00779771$

Step 11

Find the probability of $P (9)$ .

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Step 11.1

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

$p (x) = {}_{12}C_{9} \cdot p^{x} \cdot q^{n - x}$

Step 11.2

Find the value of ${}_{12}C_{9}$ .

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Step 11.2.1

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

${}_{12}C_{9} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 11.2.2

Fill in the known values.

$\frac{(12)!}{(9)! (12 - 9)!}$

Step 11.2.3

Simplify.

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Step 11.2.3.1

Subtract $9$ from $12$ .

$\frac{(12)!}{(9)! (3)!}$

Step 11.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10 \cdot 9!$ .

$\frac{12 \cdot 11 \cdot 10 \cdot 9!}{(9)! (3)!}$

Step 11.2.3.3

Cancel the common factor of $9!$ .

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Step 11.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10 \cdot 9!}{(9)! (3)!}$

Step 11.2.3.3.2

Rewrite the expression.

$\frac{12 \cdot 11 \cdot 10}{(3)!}$

Step 11.2.3.4

Simplify the numerator.

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Step 11.2.3.4.1

Multiply $12$ by $11$ .

$\frac{132 \cdot 10}{(3)!}$

Step 11.2.3.4.2

Multiply $132$ by $10$ .

$\frac{1320}{(3)!}$

Step 11.2.3.5

Simplify the denominator.

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Step 11.2.3.5.1

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

$\frac{1320}{3 \cdot 2 \cdot 1}$

Step 11.2.3.5.2

Multiply $3 \cdot 2 \cdot 1$ .

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Step 11.2.3.5.2.1

Multiply $3$ by $2$ .

$\frac{1320}{6 \cdot 1}$

Step 11.2.3.5.2.2

Multiply $6$ by $1$ .

$\frac{1320}{6}$

Step 11.2.3.6

Divide $1320$ by $6$ .

$220$

Step 11.3

Fill the known values into the equation.

$220 \cdot {(0.3)}^{9} \cdot {(1 - 0.3)}^{12 - 9}$

Step 11.4

Simplify the result.

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Step 11.4.1

Raise $0.3$ to the power of $9$ .

$220 \cdot 0.00001968 \cdot {(1 - 0.3)}^{12 - 9}$

Step 11.4.2

Multiply $220$ by $0.00001968$ .

$0.00433026 \cdot {(1 - 0.3)}^{12 - 9}$

Step 11.4.3

Subtract $0.3$ from $1$ .

$0.00433026 \cdot {0.7}^{12 - 9}$

Step 11.4.4

Subtract $9$ from $12$ .

$0.00433026 \cdot {0.7}^{3}$

Step 11.4.5

Raise $0.7$ to the power of $3$ .

$0.00433026 \cdot 0.343$

Step 11.4.6

Multiply $0.00433026$ by $0.343$ .

$0.00148527$

Step 12

Find the probability of $P (10)$ .

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Step 12.1

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

$p (x) = {}_{12}C_{10} \cdot p^{x} \cdot q^{n - x}$

Step 12.2

Find the value of ${}_{12}C_{10}$ .

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Step 12.2.1

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

${}_{12}C_{10} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 12.2.2

Fill in the known values.

$\frac{(12)!}{(10)! (12 - 10)!}$

Step 12.2.3

Simplify.

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Step 12.2.3.1

Subtract $10$ from $12$ .

$\frac{(12)!}{(10)! (2)!}$

Step 12.2.3.2

Rewrite $(12)!$ as $12 \cdot 11 \cdot 10!$ .

$\frac{12 \cdot 11 \cdot 10!}{(10)! (2)!}$

Step 12.2.3.3

Reduce the expression by cancelling the common factors.

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Step 12.2.3.3.1

Cancel the common factor of $10!$ .

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Step 12.2.3.3.1.1

Cancel the common factor.

$\frac{12 \cdot 11 \cdot 10!}{(10)! (2)!}$

Step 12.2.3.3.1.2

Rewrite the expression.

$\frac{12 \cdot 11}{(2)!}$

Step 12.2.3.3.2

Multiply $12$ by $11$ .

$\frac{132}{(2)!}$

Step 12.2.3.4

Simplify the denominator.

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Step 12.2.3.4.1

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

$\frac{132}{2 \cdot 1}$

Step 12.2.3.4.2

Multiply $2$ by $1$ .

$\frac{132}{2}$

Step 12.2.3.5

Divide $132$ by $2$ .

$66$

Step 12.3

Fill the known values into the equation.

$66 \cdot {(0.3)}^{10} \cdot {(1 - 0.3)}^{12 - 10}$

Step 12.4

Simplify the result.

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Step 12.4.1

Raise $0.3$ to the power of $10$ .

$66 \cdot 0.0000059 \cdot {(1 - 0.3)}^{12 - 10}$

Step 12.4.2

Multiply $66$ by $0.0000059$ .

$0.00038972 \cdot {(1 - 0.3)}^{12 - 10}$

Step 12.4.3

Subtract $0.3$ from $1$ .

$0.00038972 \cdot {0.7}^{12 - 10}$

Step 12.4.4

Subtract $10$ from $12$ .

$0.00038972 \cdot {0.7}^{2}$

Step 12.4.5

Raise $0.7$ to the power of $2$ .

$0.00038972 \cdot 0.49$

Step 12.4.6

Multiply $0.00038972$ by $0.49$ .

$0.00019096$

Step 13

Find the probability of $P (11)$ .

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Step 13.1

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

$p (x) = {}_{12}C_{11} \cdot p^{x} \cdot q^{n - x}$

Step 13.2

Find the value of ${}_{12}C_{11}$ .

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Step 13.2.1

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

${}_{12}C_{11} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 13.2.2

Fill in the known values.

$\frac{(12)!}{(11)! (12 - 11)!}$

Step 13.2.3

Simplify.

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Step 13.2.3.1

Subtract $11$ from $12$ .

$\frac{(12)!}{(11)! (1)!}$

Step 13.2.3.2

Rewrite $(12)!$ as $12 \cdot 11!$ .

$\frac{12 \cdot 11!}{(11)! (1)!}$

Step 13.2.3.3

Cancel the common factor of $11!$ .

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Step 13.2.3.3.1

Cancel the common factor.

$\frac{12 \cdot 11!}{(11)! (1)!}$

Step 13.2.3.3.2

Rewrite the expression.

$\frac{12}{(1)!}$

Step 13.2.3.4

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

$\frac{12}{1}$

Step 13.2.3.5

Divide $12$ by $1$ .

$12$

Step 13.3

Fill the known values into the equation.

$12 \cdot {(0.3)}^{11} \cdot {(1 - 0.3)}^{12 - 11}$

Step 13.4

Simplify the result.

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Step 13.4.1

Raise $0.3$ to the power of $11$ .

$12 \cdot 0.00000177 \cdot {(1 - 0.3)}^{12 - 11}$

Step 13.4.2

Multiply $12$ by $0.00000177$ .

$0.00002125 \cdot {(1 - 0.3)}^{12 - 11}$

Step 13.4.3

Subtract $0.3$ from $1$ .

$0.00002125 \cdot {0.7}^{12 - 11}$

Step 13.4.4

Subtract $11$ from $12$ .

$0.00002125 \cdot {0.7}^{1}$

Step 13.4.5

Evaluate the exponent.

$0.00002125 \cdot 0.7$

Step 13.4.6

Multiply $0.00002125$ by $0.7$ .

$0.00001488$

Step 14

Find the probability of $P (12)$ .

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Step 14.1

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

$p (x) = {}_{12}C_{12} \cdot p^{x} \cdot q^{n - x}$

Step 14.2

Find the value of ${}_{12}C_{12}$ .

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Step 14.2.1

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

${}_{12}C_{12} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Step 14.2.2

Fill in the known values.

$\frac{(12)!}{(12)! (12 - 12)!}$

Step 14.2.3

Simplify.

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Step 14.2.3.1

Cancel the common factor of $(12)!$ .

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Step 14.2.3.1.1

Cancel the common factor.

$\frac{(12)!}{(12)! (12 - 12)!}$

Step 14.2.3.1.2

Rewrite the expression.

$\frac{1}{(12 - 12)!}$

Step 14.2.3.2

Simplify the denominator.

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Step 14.2.3.2.1

Subtract $12$ from $12$ .

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

Step 14.2.3.2.2

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

$\frac{1}{1}$

Step 14.2.3.3

Divide $1$ by $1$ .

$1$

Step 14.3

Fill the known values into the equation.

$1 \cdot {(0.3)}^{12} \cdot {(1 - 0.3)}^{12 - 12}$

Step 14.4

Simplify the result.

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Step 14.4.1

Multiply ${(0.3)}^{12}$ by $1$ .

${(0.3)}^{12} \cdot {(1 - 0.3)}^{12 - 12}$

Step 14.4.2

Raise $0.3$ to the power of $12$ .

$0.00000053 \cdot {(1 - 0.3)}^{12 - 12}$

Step 14.4.3

Subtract $0.3$ from $1$ .

$0.00000053 \cdot {0.7}^{12 - 12}$

Step 14.4.4

Subtract $12$ from $12$ .

$0.00000053 \cdot {0.7}^{0}$

Step 14.4.5

Anything raised to $0$ is $1$ .

$0.00000053 \cdot 1$

Step 14.4.6

Multiply $0.00000053$ by $1$ .

$0.00000053$

Step 15

The probability $P (x > 0)$ is the sum of the probabilities of all possible $x$ values between $0$ and $n$ . $P (x > 0) = P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6) + P (x = 7) + P (x = 8) + P (x = 9) + P (x = 10) + P (x = 11) + P (x = 12) = 0.07118376 + 0.16779029 + 0.23970042 + 0.23113969 + 0.15849579 + 0.07924789 + 0.02911147 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053 = 0.98615871$ .

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Step 15.1

Add $0.07118376$ and $0.16779029$ .

$p (x > 0) = 0.23897406 + 0.23970042 + 0.23113969 + 0.15849579 + 0.07924789 + 0.02911147 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.2

Add $0.23897406$ and $0.23970042$ .

$p (x > 0) = 0.47867448 + 0.23113969 + 0.15849579 + 0.07924789 + 0.02911147 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.3

Add $0.47867448$ and $0.23113969$ .

$p (x > 0) = 0.70981418 + 0.15849579 + 0.07924789 + 0.02911147 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.4

Add $0.70981418$ and $0.15849579$ .

$p (x > 0) = 0.86830997 + 0.07924789 + 0.02911147 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.5

Add $0.86830997$ and $0.07924789$ .

$p (x > 0) = 0.94755786 + 0.02911147 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.6

Add $0.94755786$ and $0.02911147$ .

$p (x > 0) = 0.97666934 + 0.00779771 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.7

Add $0.97666934$ and $0.00779771$ .

$p (x > 0) = 0.98446705 + 0.00148527 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.8

Add $0.98446705$ and $0.00148527$ .

$p (x > 0) = 0.98595233 + 0.00019096 + 0.00001488 + 0.00000053$

Step 15.9

Add $0.98595233$ and $0.00019096$ .

$p (x > 0) = 0.9861433 + 0.00001488 + 0.00000053$

Step 15.10

Add $0.9861433$ and $0.00001488$ .

$p (x > 0) = 0.98615818 + 0.00000053$

Step 15.11

Add $0.98615818$ and $0.00000053$ .

$p (x > 0) = 0.98615871$