# Solved:Conservation staff running a rare-bird breeding program are monitoring the progress of 15 eggs. Each egg has a probability

#### ByDr. Raju Chaudhari

Sep 26, 2020

Conservation staff running a rare-bird breeding program are monitoring the progress of 15 eggs. Each egg has a probability of 0.8 of successfully hatching. Let X be the number of eggs that
hatch successfully.

i) Assuming that the 15 egg hatching are independent events, what type of distribution describes the random variable X?
ii) What is the expected number of eggs that will hatch?
iii) What is the probability that at least 11 eggs will hatch successfully?

#### Solution

Let $X$ denote the number of eggs that hatch successfully out of $n=15$ eggs. Each egg has a probability of $p=0.8$ of successfully hatching.

i) Assuming that the 15 egg hatching are independent events, the random variable $X$ describes Binomial distribution with $n=15$ and $p= 0.8$.

That is $X\sim B(15, 0.8)$.

The probability mass function of $X$ is

$$P(X=x) = \binom{15}{x} (0.8)^x (1-0.8)^{15-x}, \; x=0,1,\cdots, 15.$$

ii) The expected number of eggs that will hatch is

 \begin{aligned} E(X) &= np\\ &= 15 \times 0.8\\ &= 12 \end{aligned}

iii) The probability that at least 11 eggs will hatch successfully is

 \begin{aligned} P(X\geq 11) &= \sum_{x=11}^{15} P(X=x)\\ &= \bigg[P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)\bigg]\\ &= \binom{15}{11} (0.8)^{11} (1-0.8)^{15-11} +\binom{15}{12} (0.8)^{12} (1-0.8)^{15-12}\\ &\quad + \binom{15}{13} (0.8)^{13} (1-0.8)^{15-13}+\binom{15}{14} (0.8)^{14} (1-0.8)^{15-14}\\ &\quad + \binom{15}{15} (0.8)^{15} (1-0.8)^{15-15}\\ &=0.1876+0.2501+0.2309+0.1319+0.0352\\ & = 0.8358 \end{aligned}