Question

22​% adults favor the use of unmanned drones by police agencies. Twelve U.S. adults are randomly selected. Find the probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is​ (a) exactly​ three, (b) at least​ four, (c) less than eight. ​(a) Upper P left parenthesis 3 right parenthesis equals nothing ​(Round to three decimal places as​ needed.)

Answer 1

is the probability of an adult favouring use of unmanned drones

p = 22% = 0.22

Number of points in the sample is $n=12$

The distribution of random variable which is the number of people out of $n=12$ who favour use of unmanned drones by police agencies where
p = 22% = 0.22
is the probability of any one adult favouring use of unmanned drones is a Binomial distribution with parameters

$X=\text{Binom}(n,p)$ with $n=12,\,p=0.22$

We have $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$

a) $P(X=3)=\binom{12}{3}(0.22)^3(1-0.22)^{12-3}\approx {\color{Red} 0.25}$ is the probability that exactly 3 adults of the sample of 12 people favour use of unmanned drones

b) $P(X\geq 4)=P(X=4)+P(X=5)+\cdots P(X=12)$

Which is $\binom{12}{3}(0.22)^3(1-0.22)^{12-3}+\binom{12}{4}(0.22)^4(1-0.22)^{12-4}+\cdots \binom{12}{12}(0.22)^{12}(1-0.22)^{12-12}$

Which equals $P(X\geq 4)\approx {\color{Red} 0.261}$

c) $P(X\leq 8)=P(X=8)+P(X=7)+\cdots P(X=0)$ which equals

$\binom{12}{8}(0.22)^8(1-0.22)^{12-8}+\binom{12}{7}(0.22)^7(1-0.22)^{12-7}+\cdots \binom{12}{0}(0.22)^{0}(1-0.22)^{12-0}$

Which is (Which actually becomes 1 when rounding to 3 decimal places so here I have rounded to 5 decimal places)

P(X 8)0.99986

$\blacksquare$

Please do rate this answer positively if you found it helpful. Thanks and have a good day!