Question

79% of all bald eagles survive their first year of life. give your answers as decimals, not percents. if 48 bald eagles are randomly selected, find the probability that

a. exactly 38 of them survive their first year of life.

b. at most 36 of them survive their first year of life.

c. at least 39 of them survive their first year of life.

d. between 35 and 43 (including 35 and 43) of them survive their first year of life.

Explanation:

Step1: Identify binomial parameters

Let $n = 48$ (number of bald - eagles selected), $p=0.79$ (probability of a bald - eagle surviving its first year), and $q = 1 - p=1 - 0.79 = 0.21$. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate probability for part a

For $k = 38$, $C(48,38)=\frac{48!}{38!(48 - 38)!}=\frac{48!}{38!10!}=\frac{48\times47\times\cdots\times39}{10!}$
$P(X = 38)=C(48,38)\times(0.79)^{38}\times(0.21)^{10}$
$C(48,38)=\frac{48!}{38!10!}=\frac{48!}{38!10!}=6540715896$
$P(X = 38)=6540715896\times(0.79)^{38}\times(0.21)^{10}\approx0.137$

Step3: Calculate probability for part b

$P(X\leq16)=\sum_{k = 0}^{16}C(48,k)\times(0.79)^{k}\times(0.21)^{48 - k}$. Using a binomial probability calculator or software (e.g., in R: pbinom(16,48,0.79)), we get $P(X\leq16)\approx0$

Step4: Calculate probability for part c

$P(X\geq19)=1 - P(X\lt19)=1-\sum_{k = 0}^{18}C(48,k)\times(0.79)^{k}\times(0.21)^{48 - k}$
Using a binomial probability calculator or software (e.g., in R: 1 - pbinom(18,48,0.79)), we get $P(X\geq19)\approx1$

Step5: Calculate probability for part d

$P(35\leq X\leq43)=\sum_{k = 35}^{43}C(48,k)\times(0.79)^{k}\times(0.21)^{48 - k}$
Using a binomial probability calculator or software (e.g., in R: pbinom(43,48,0.79)-pbinom(34,48,0.79)), we get $P(35\leq X\leq43)\approx0.874$

Answer:

a. $0.137$
b. $0$
c. $1$
d. $0.874$