Overbooking flight

A small regional carrier accepted 12 reservations for a particular flight with 11 seats. Seven reservations went to regular customers who would arrive for the flight. Each remaining passenger will arrive for the flight with a 49% chance, independently of each other.
(Report answers accurate up to 4 decimal places. )

A. Find the probability that overbooking occurs.
B. Find the probability that the flight has empty seats.

Final Answer:

o =  0.0282
e =  0.8248

Step-by-step explanation:

$$\begin{gathered} p=49\%=\frac{49}{100}=0.49 \\ r=7 \\ n=12-r=12-7=5 \\ s=11-r=11-7=4 \\ s=4seats \\ overbook:5arrive \\ OK:4arrive \\ free:0,1,2,3arrive \\ o=\left(\genfrac{}{}{0ex}{}{n}{5}\right)\cdot p^5\cdot (1-p{)}^{n-5}=(\genfrac{}{}{0ex}{}{5}{5})\cdot 0.49^5\cdot (1-0.49{)}^{5-5}=1\cdot 0.49^5\cdot 0.51^0=0.0282 \end{gathered}$$

$$\begin{gathered} e_1=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{5}{3})\cdot 0.49^3\cdot (1-0.49{)}^{5-3}=10\cdot 0.49^3\cdot 0.51^2\doteq 0.306 \\ {e}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{5}{2})\cdot 0.49^2\cdot (1-0.49{)}^{5-2}=10\cdot 0.49^2\cdot 0.51^3\doteq 0.3185 \\ {e}_3=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{5}{1})\cdot 0.49^1\cdot (1-0.49{)}^{5-1}=5\cdot 0.49^1\cdot 0.51^4\doteq 0.1657 \\ {e}_4=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{5}{0})\cdot 0.49^0\cdot (1-0.49{)}^{5-0}=1\cdot 0.49^0\cdot 0.51^5\doteq 0.0345 \\ e=e_1+e_2+e_3+e_4=0.306+0.3185+0.1657+0.0345=0.8248 \end{gathered}$$

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