suppose that the probability is 0.0122 that a person who purchases a ticket for an airplane does not show up for the flight. if 250 tickets are sold for an airplane flight that has 247 seats, what is the probability that there will not be enough seats available for this flight?
a. 0.4107
b. 0.1909
c. 0.5893
d. 0.6360

Question

Accepted Answer

# Explanation:
## Step1: Define the binomial distribution
Let $X$ be the number of people who do not show - up. $n = 250$ (number of tickets sold) and $p=0.0122$. We want to find $P(X < 250 - 247)=P(X < 3)$.
## Step2: Use the binomial probability formula
The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.
We need to calculate $P(X = 0)+P(X = 1)+P(X = 2)$.
For $k = 0$:
$$
\begin{align*}
P(X = 0)&=C(250,0)\times(0.0122)^{0}\times(1 - 0.0122)^{250-0}\
&=1\times1\times(0.9878)^{250}\
\end{align*}
$$
For $k = 1$:
$$
\begin{align*}
P(X = 1)&=C(250,1)\times(0.0122)^{1}\times(1 - 0.0122)^{250 - 1}\
&=250\times0.0122\times(0.9878)^{249}
\end{align*}
$$
For $k = 2$:
$$
\begin{align*}
P(X = 2)&=C(250,2)\times(0.0122)^{2}\times(1 - 0.0122)^{250-2}\
&=\frac{250!}{2!(250 - 2)!}\times(0.0122)^{2}\times(0.9878)^{248}\
&=\frac{250\times249}{2}\times(0.0122)^{2}\times(0.9878)^{248}
\end{align*}
$$
## Step3: Calculate the sum
$$
\begin{align*}
P(X<3)&=P(X = 0)+P(X = 1)+P(X = 2)\
&\approx0.5893
\end{align*}
$$
# Answer:
C. 0.5893

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QUESTION IMAGE

Question

Explanation:

Step1: Define the binomial distribution

Step2: Use the binomial probability formula

Step3: Calculate the sum

Answer: