Question

a local county has an unemployment rate of 4%. a random sample of 30 employable people are picked at random from the county and are asked if they are employed. round answers to 4 decimal places.
a) find the probability that exactly 3 in the sample are unemployed.
b) find the probability that there are fewer than 3 in the sample are unemployed.
c) find the probability that there are more than 1 in the sample are unemployed.
d) find the probability that there are at most 3 in the sample are unemployed.

Explanation:

Step1: Identify the binomial - distribution parameters

The binomial - distribution formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 30$, $p=0.04$, and $1 - p = 0.96$.

Step2: Calculate the probability for part a

For $k = 3$:
$C(30,3)=\frac{30!}{3!(30 - 3)!}=\frac{30\times29\times28}{3\times2\times1}=4060$
$P(X = 3)=C(30,3)\times(0.04)^{3}\times(0.96)^{27}$
$P(X = 3)=4060\times0.04^{3}\times0.96^{27}\approx4060\times0.000064\times0.3377\approx0.0878$

Step3: Calculate the probability for part b

$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$
$P(X = 0)=C(30,0)\times(0.04)^{0}\times(0.96)^{30}=1\times1\times0.2938\approx0.2938$
$C(30,1)=\frac{30!}{1!(30 - 1)!}=30$
$P(X = 1)=C(30,1)\times(0.04)^{1}\times(0.96)^{29}=30\times0.04\times0.3060\approx0.3672$
$C(30,2)=\frac{30!}{2!(30 - 2)!}=\frac{30\times29}{2\times1}=435$
$P(X = 2)=C(30,2)\times(0.04)^{2}\times(0.96)^{28}=435\times0.0016\times0.3187\approx0.2215$
$P(X\lt3)=0.2938 + 0.3672+0.2215 = 0.8825$

Step4: Calculate the probability for part c

$P(X\gt1)=1-(P(X = 0)+P(X = 1))$
$P(X = 0)+P(X = 1)=0.2938 + 0.3672=0.6610$
$P(X\gt1)=1 - 0.6610=0.3390$

Step5: Calculate the probability for part d

$P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)$
$P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)=0.2938+0.3672 + 0.2215+0.0878=0.9703$

Answer:

a) $0.0878$
b) $0.8825$
c) $0.3390$
d) $0.9703$