Question

in this problem, round to four decimals when possible.
15% of all americans live in poverty. if 48 americans are randomly selected, find the probability that
a. exactly 7 of them live in poverty.
b. at most 6 of them live in poverty.
c. at least 5 of them live in poverty.
d. between 4 and 9 (including 4 and 9) of them live in poverty.
hint:
hint
video on finding binomial probabilities +

Explanation:

This is a binomial probability problem. The binomial probability 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, $n = 48$, $p=0.15$, and $1 - p = 0.85$.

Step1: Calculate combination for part a

Calculate $C(48,7)=\frac{48!}{7!(48 - 7)!}=\frac{48!}{7!41!}=\frac{48\times47\times46\times45\times44\times43\times42}{7\times6\times5\times4\times3\times2\times1}=73629072$.
$P(X = 7)=C(48,7)\times(0.15)^{7}\times(0.85)^{41}$
$P(X = 7)=73629072\times(0.15)^{7}\times(0.85)^{41}\approx0.1462$

Step2: Calculate cumulative probability for part b

$P(X\leq6)=\sum_{k = 0}^{6}C(48,k)\times(0.15)^{k}\times(0.85)^{48 - k}$
$P(X = 0)=C(48,0)\times(0.15)^{0}\times(0.85)^{48}=(0.85)^{48}\approx0.0004$
$P(X = 1)=C(48,1)\times(0.15)^{1}\times(0.85)^{47}=48\times0.15\times(0.85)^{47}\approx0.0030$
$P(X = 2)=C(48,2)\times(0.15)^{2}\times(0.85)^{46}=\frac{48!}{2!46!}\times(0.15)^{2}\times(0.85)^{46}\approx0.0107$
$P(X = 3)=C(48,3)\times(0.15)^{3}\times(0.85)^{45}=\frac{48!}{3!45!}\times(0.15)^{3}\times(0.85)^{45}\approx0.0262$
$P(X = 4)=C(48,4)\times(0.15)^{4}\times(0.85)^{44}=\frac{48!}{4!44!}\times(0.15)^{4}\times(0.85)^{44}\approx0.0492$
$P(X = 5)=C(48,5)\times(0.15)^{5}\times(0.85)^{43}=\frac{48!}{5!43!}\times(0.15)^{5}\times(0.85)^{43}\approx0.0777$
$P(X = 6)=C(48,6)\times(0.15)^{6}\times(0.85)^{42}=\frac{48!}{6!42!}\times(0.15)^{6}\times(0.85)^{42}\approx0.1048$
$P(X\leq6)\approx0.0004 + 0.0030+0.0107 + 0.0262+0.0492+0.0777+0.1048=0.2720$

Step3: Calculate complementary probability for part c

$P(X\geq5)=1 - P(X\leq4)$
$P(X\leq4)=\sum_{k = 0}^{4}C(48,k)\times(0.15)^{k}\times(0.85)^{48 - k}$
$P(X\leq4)\approx0.0004+0.0030 + 0.0107+0.0262+0.0492=0.0895$
$P(X\geq5)=1 - 0.0895 = 0.9105$

Step4: Calculate cumulative probabilities for part d

$P(4\leq X\leq9)=\sum_{k = 4}^{9}C(48,k)\times(0.15)^{k}\times(0.85)^{48 - k}$
$P(4\leq X\leq9)=P(X\leq9)-P(X\leq3)$
$P(X\leq9)$: Calculate $P(X = 7)$ (done in part a), $P(X = 8)=C(48,8)\times(0.15)^{8}\times(0.85)^{40}$, $P(X = 9)=C(48,9)\times(0.15)^{9}\times(0.85)^{39}$ and sum with $P(X\leq6)$
$P(X\leq9)\approx0.7574$
$P(X\leq3)\approx0.0403$
$P(4\leq X\leq9)=0.7574 - 0.0403=0.7171$

Answer:

a. $0.1462$
b. $0.2720$
c. $0.9105$
d. $0.7171$