Finite Math Examples

$x = 2$ , $n = 4$ , $p = 0.6$

Step 1

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

$p (x) = {}_{4}C_{2} \cdot p^{x} \cdot q^{n - x}$

Step 2

Find the value of ${}_{4}C_{2}$ .

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

${}_{4}C_{2} = {}_{n}C_{r} = \frac{n!}{(r)! (n - r)!}$

Fill in the known values.

$\frac{(4)!}{(2)! (4 - 2)!}$

Simplify.

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Subtract $2$ from $4$ .

$\frac{(4)!}{(2)! (2)!}$

Rewrite $(4)!$ as $4 \cdot 3 \cdot 2!$ .

$\frac{4 \cdot 3 \cdot 2!}{(2)! (2)!}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor of $2!$ .

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Cancel the common factor.

$\frac{4 \cdot 3 \cdot 2!}{(2)! (2)!}$

Rewrite the expression.

$\frac{4 \cdot 3}{(2)!}$

Multiply $4$ by $3$ .

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

Simplify the denominator.

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

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

Multiply $2$ by $1$ .

$\frac{12}{2}$

Divide $12$ by $2$ .

$6$

Step 3

Fill the known values into the equation.

$6 \cdot {(0.6)}^{2} \cdot {(1 - 0.6)}^{4 - 2}$

Step 4

Simplify the result.

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Raise $0.6$ to the power of $2$ .

$6 \cdot 0.36 \cdot {(1 - 0.6)}^{4 - 2}$

Multiply $6$ by $0.36$ .

$2.16 \cdot {(1 - 0.6)}^{4 - 2}$

Subtract $0.6$ from $1$ .

$2.16 \cdot {0.4}^{4 - 2}$

Subtract $2$ from $4$ .

$2.16 \cdot {0.4}^{2}$

Raise $0.4$ to the power of $2$ .

$2.16 \cdot 0.16$

Multiply $2.16$ by $0.16$ .

$0.3456$

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