+0

Binomial Distribution

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58
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A small regional carrier accepted 14 reservations for a particular flight with 12 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 45% chance, independently of each other.
(Report answers accurate to 4 decimal places.)

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

Oct 31, 2020

#1
+111600
+1

11 definites, 3 uncertain.  only 1 seat available.

P(a person will show)=0.45

P(a person won't show)=0.55

$$\text{P(no one shows)} \qquad \qquad= \binom{3}{0} * 0.45^0 * 0.55^3\\ \text{P(only one person shows)} = \binom{3}{1} * 0.45^1 * 0.55^2\\ etc$$

You should be able to take it from there.

Oct 31, 2020