Finite Math Examples

$x > 5$ , $n = 7$ , $p = 0.7$

Subtract $0.7$ from $1$ to get $0.3$ .

$0.3$

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, $p (x > 5) = P (x = 6) + P (x = 7)$ .

$p (x > 5) = P (x = 6) + P (x = 7)$

Find the probability of $P (6)$ .

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

$p (x) = {}_{7}C_{6} \cdot p^{x} \cdot q^{n - x}$

Find the value of ${}_{7}C_{6}$ .

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

${}_{7}C_{6} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Fill in the known values.

$\frac{(7)!}{(6)! (7 - 6)!}$

Simplify.

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Subtract $6$ from $7$ to get $1$ .

$\frac{(7)!}{(6)! 1!}$

Rewrite $(7)!$ as $7 \cdot 6!$ .

$\frac{7 \cdot 6!}{6! 1!}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{7 \cdot 6!}{6! 1!}$

Rewrite the expression.

$\frac{7}{1!}$

Expand $1!$ to $1$ .

$\frac{7}{1}$

Divide $7$ by $1$ to get $7$ .

$7$

Fill the known values into the equation.

$7 \cdot {(0.7)}^{6} \cdot {(1 - 0.7)}^{7 - 6}$

Simplify the result.

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Remove parentheses around $0.7$ .

$7 \cdot {0.7}^{6} \cdot {(1 - 0.7)}^{7 - 6}$

Raise $0.7$ to the power of $6$ to get $0.117649$ .

$7 \cdot 0.117649 \cdot {(1 - 0.7)}^{7 - 6}$

Multiply $7$ by $0.117649$ to get $0.823543$ .

$0.823543 \cdot {(1 - 0.7)}^{7 - 6}$

Subtract $0.7$ from $1$ to get $0.3$ .

$0.823543 \cdot {(0.3)}^{7 - 6}$

Subtract $6$ from $7$ to get $1$ .

$0.823543 \cdot {(0.3)}^{1}$

Remove parentheses around $0.3$ .

$0.823543 \cdot {0.3}^{1}$

Evaluate the exponent.

$0.823543 \cdot 0.3$

Multiply $0.823543$ by $0.3$ to get $0.2470629$ .

$0.2470629$

Find the probability of $P (7)$ .

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Find the probability of $P (6)$ .

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

$p (x) = {}_{7}C_{6} \cdot p^{x} \cdot q^{n - x}$

Find the value of ${}_{7}C_{6}$ .

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

${}_{7}C_{6} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Fill in the known values.

$\frac{(7)!}{(6)! (7 - 6)!}$

Simplify.

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Subtract $6$ from $7$ to get $1$ .

$\frac{(7)!}{(6)! 1!}$

Rewrite $(7)!$ as $7 \cdot 6!$ .

$\frac{7 \cdot 6!}{6! 1!}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{7 \cdot 6!}{6! 1!}$

Rewrite the expression.

$\frac{7}{1!}$

Expand $1!$ to $1$ .

$\frac{7}{1}$

Divide $7$ by $1$ to get $7$ .

$7$

Fill the known values into the equation.

$7 \cdot {(0.7)}^{6} \cdot {(1 - 0.7)}^{7 - 6}$

Simplify the result.

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Remove parentheses around $0.7$ .

$7 \cdot {0.7}^{6} \cdot {(1 - 0.7)}^{7 - 6}$

Raise $0.7$ to the power of $6$ to get $0.117649$ .

$7 \cdot 0.117649 \cdot {(1 - 0.7)}^{7 - 6}$

Multiply $7$ by $0.117649$ to get $0.823543$ .

$0.823543 \cdot {(1 - 0.7)}^{7 - 6}$

Subtract $0.7$ from $1$ to get $0.3$ .

$0.823543 \cdot {(0.3)}^{7 - 6}$

Subtract $6$ from $7$ to get $1$ .

$0.823543 \cdot {(0.3)}^{1}$

Remove parentheses around $0.3$ .

$0.823543 \cdot {0.3}^{1}$

Evaluate the exponent.

$0.823543 \cdot 0.3$

Multiply $0.823543$ by $0.3$ to get $0.2470629$ .

$0.2470629$

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

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

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

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

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

Fill in the known values.

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

Simplify.

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Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{(7 - 7)!}$

Simplify the denominator.

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Subtract $7$ from $7$ to get $0$ .

$\frac{1}{0!}$

Expand $0!$ to $1$ .

$\frac{1}{1}$

Divide $1$ by $1$ to get $1$ .

$1$

Fill the known values into the equation.

$1 \cdot {(0.7)}^{7} \cdot {(1 - 0.7)}^{7 - 7}$

Simplify the result.

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Multiply ${(0.7)}^{7}$ by $1$ to get ${(0.7)}^{7}$ .

${(0.7)}^{7} \cdot {(1 - 0.7)}^{7 - 7}$

Remove parentheses around $0.7$ .

${0.7}^{7} \cdot {(1 - 0.7)}^{7 - 7}$

Raise $0.7$ to the power of $7$ to get $0.0823543$ .

$0.0823543 \cdot {(1 - 0.7)}^{7 - 7}$

Subtract $0.7$ from $1$ to get $0.3$ .

$0.0823543 \cdot {(0.3)}^{7 - 7}$

Subtract $7$ from $7$ to get $0$ .

$0.0823543 \cdot {(0.3)}^{0}$

Remove parentheses around $0.3$ .

$0.0823543 \cdot {0.3}^{0}$

Anything raised to $0$ is $1$ .

$0.0823543 \cdot 1$

Multiply $0.0823543$ by $1$ to get $0.0823543$ .

$0.0823543$

Add $0.2470629$ and $0.0823543$ to get $0.3294172$ .

$p (x > 5) = 0.3294172$

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