3) (4 points) suppose that 10 percent of the tubes produced by a machine are defective. if 6 tubes are inspected at random, determine the probability that three tubes are defective.

Question

Accepted Answer

# Explanation:
## Step1: Identify the formula
Use the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 6$ (number of trials, i.e., tubes inspected), $k = 3$ (number of "successes", i.e., defective tubes), $p=0.1$ (probability of a tube being defective), and $C(n,k)=\frac{n!}{k!(n - k)!}$

## Step2: Calculate the combination $C(n,k)$
$$
\begin{align*}
C(6,3)&=\frac{6!}{3!(6 - 3)!}\
&=\frac{6!}{3!3!}\
&=\frac{6\times5\times4\times3!}{3!\times3\times2\times 1}\
&=20
\end{align*}
$$

## Step3: Calculate $p^{k}$ and $(1 - p)^{n - k}$
$p^{k}=(0.1)^{3}=0.001$, $(1 - p)^{n - k}=(0.9)^{3}=0.729$

## Step4: Calculate the probability $P(X = 3)$
$$
\begin{align*}
P(X = 3)&=C(6,3)\times p^{3}\times(0.9)^{3}\
&=20\times0.001\times0.729\
&=0.01458
\end{align*}
$$

# Answer:
$0.01458$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the formula

Step2: Calculate the combination \(C(n,k)\)

Step3: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

Step4: Calculate the probability \(P(X = 3)\)

Answer: