Question

assume that a procedure yields a binomial distribution with a trial repeated $n$ times. use the binomial probability formula to find the probability of $x$ successes given the probability $p$ of success on a single trial. round to three decimal places.
$n = 10, x = 4, p = 0.25$
0.179
0.125
0.146

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is:
$$P(x) = \binom{n}{x} p^x (1-p)^{n-x}$$
where $\binom{n}{x} = \frac{n!}{x!(n-x)!}$

Step2: Calculate combination term

For $n=10, x=4$:
$$\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10\times9\times8\times7}{4\times3\times2\times1} = 210$$

Step3: Compute $p^x$ and $(1-p)^{n-x}$

$p=0.25$, so:
$$p^x = 0.25^4 = 0.00390625$$
$$(1-p)^{n-x} = (0.75)^{6} = 0.177978515625$$

Step4: Multiply all terms

$$P(4) = 210 \times 0.00390625 \times 0.177978515625$$
$$P(4) = 210 \times 0.000694580078125 = 0.14586181640625$$

Step5: Round to 3 decimal places

Round $0.14586181640625$ to three decimal places: $0.146$

Answer:

0.146