Question

21​% 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

Answer 1

Answer:

a) $P(X=3)=(12C3)(0.21)^3 (1-0.21)^{12-3}=0.2442$

b) $P(X \geq 4) =1- P(X<4) =1-P(X\leq 3)$

And we can find the individual probabilities:

$P(X=0)=(12C0)(0.21)^0 (1-0.21)^{12-0}=0.059$

$P(X=1)=(12C1)(0.21)^1 (1-0.21)^{12-1}=0.1885$

$P(X=2)=(12C2)(0.21)^2 (1-0.21)^{12-2}=0.2756$

$P(X=3)=(12C3)(0.21)^3 (1-0.21)^{12-3}=0.2442$

And replacing we got:

$P(X \geq 4)= 1-[P(X=0)+P(X=1) +P(X=2) +P(X=3)]= 1 -[0.059+0.1885+0.2756+0.2442]= 0.2327$

c) $P(X=8)=(12C8)(0.21)^8 (1-0.21)^{12-8}=0.000729$

$P(X=9)=(12C9)(0.21)^9 (1-0.21)^{12-9}=8.62x10^{-5}$

$P(X=10)=(12C10)(0.21)^{10} (1-0.21)^{12-10}=6.87x10^{-6}$

$P(X=11)=(12C11)(0.21)^{11} (1-0.21)^{12-11}=3.32x10^{-7}$

$P(X=12)=(12C12)(0.21)^{12} (1-0.21)^{12-12}=7.36x10^{-9}$

And replacing we got:

$P(X<8) = 0.94$

Step-by-step explanation:

Let X the random variable of interest "number of US adults who favor the use of unmanned drones", on this case we now that:

$X \sim Binom(n=12, p=0.21)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=3)$

$P(X=3)=(12C3)(0.21)^3 (1-0.21)^{12-3}=0.2442$

Part b

We want to find this probability:

$P(X \geq 4)$

And we can use the complement rule and we got:

$P(X \geq 4) =1- P(X<4) =1-P(X\leq 3)$

And we can find the individual probabilities:

$P(X=0)=(12C0)(0.21)^0 (1-0.21)^{12-0}=0.059$

$P(X=1)=(12C1)(0.21)^1 (1-0.21)^{12-1}=0.1885$

$P(X=2)=(12C2)(0.21)^2 (1-0.21)^{12-2}=0.2756$

$P(X=3)=(12C3)(0.21)^3 (1-0.21)^{12-3}=0.2442$

And replacing we got:

$P(X \geq 4)= 1-[P(X=0)+P(X=1) +P(X=2) +P(X=3)]= 1-[0.059+0.1885+0.2756+0.2442]= 0.2327$

Part c

We want this probability:

$P(X<8)$

And using the complement rule got:

$P(X<8) =1-P(X \geq 8)= 1- [P(X=8) +P(X=9) +P(X=10) +P(X=11)+P(X=12)]$

And we have this:

$P(X=8)=(12C8)(0.21)^8 (1-0.21)^{12-8}=0.000729$

$P(X=9)=(12C9)(0.21)^9 (1-0.21)^{12-9}=8.62x10^{-5}$

$P(X=10)=(12C10)(0.21)^{10} (1-0.21)^{12-10}=6.87x10^{-6}$

$P(X=11)=(12C11)(0.21)^{11} (1-0.21)^{12-11}=3.32x10^{-7}$

$P(X=12)=(12C12)(0.21)^{12} (1-0.21)^{12-12}=7.36x10^{-9}$

And replacing we got:

$P(X<8) = 0.94$