from experience, an airline knows that only 73% of the passengers booked on a flight from new york to los…

from experience, an airline knows that only 73% of the passengers booked on a flight from new york to los angeles actually board their flight. if this percentage is correct, what is the probability that, in a random sample of 6 booked passengers from new york to los angeles, exactly 3 show up? round your answer to three decimal places.

from experience, an airline knows that only 73% of the passengers booked on a flight from new york to los angeles actually board their flight. if this percentage is correct, what is the probability that, in a random sample of 6 booked passengers from new york to los angeles, exactly 3 show up? round your answer to three decimal places.

Answer

Explanation:

Step1: Identify the binomial - probability formula

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 successful trials, $p$ is the probability of success in a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine the values of $n$, $k$, and $p$

Here, $n = 6$ (the number of booked passengers), $k = 3$ (the number of passengers who show up), and $p=0.73$ (the probability that a single - booked passenger shows up), and $1 - p = 1-0.73 = 0.27$.

Step3: Calculate the combination $C(n,k)$

$C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4\times3!}{3!\times3\times2\times1}=\frac{6\times5\times4}{3\times2\times1}=20$.

Step4: Calculate the probability $P(X = 3)$

$P(X = 3)=C(6,3)\times p^{3}\times(1 - p)^{6 - 3}=20\times(0.73)^{3}\times(0.27)^{3}$. $(0.73)^{3}=0.73\times0.73\times0.73 = 0.389017$, $(0.27)^{3}=0.27\times0.27\times0.27 = 0.019683$. $P(X = 3)=20\times0.389017\times0.019683\approx0.153$.

Answer:

$0.153$