Chapter 13.6, Problem 39E

To determine

To find: the probability that every person who shows up for the flight will get seat.

Answer to Problem 39E

The probability of every person who shows up for the flight there will be seat is $0.807$ .

Explanation of Solution

Given:

  $P\left(\text{not showing up}\right)=4\%=\frac{4}{100}.$ .

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

According to the given:

  $\begin{array}{l}P\left(\text{not showing up}\right)=4\%=\frac{4}{100}.\\ P\left(\text{showing up}\right)=1-\frac{4}{100}=\frac{96}{100}.\end{array}$

Out of $75$ seats sell that has $73$ seats over booking.

The probability of every person who shows up for the flight there will be seat or the probability of at least $2$ peoples do not show up is:

  $\begin{array}{l}P\left(\text{atleast }2\text{ people do not show up}\right)=1-\left(P\left(0\text{}\text{ people do not show up}\right)+P\left(1\text{}\text{ people do not show up}\right).\right)\\ \Rightarrow 1-\left(C{}_0\times {\left(\frac{4}{100}\right)}^0\times {\left(\frac{96}{100}\right)}^{75}+C{}_1\times {\left(\frac{4}{100}\right)}^1\times {\left(\frac{96}{100}\right)}^{74}\right).\\ \Rightarrow \mathrm{0.807.}\end{array}$

Hence, the probability of every person who shows up for the flight there will be seat is $0.807$ .