What is the probability of seat available

Question

An airplane knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up?

Answer 1

1 - (Probability of 51 people coming + Probability of 52 people coming)

$= 1 - \left(0.95^{51} \times 0.05 \times {}^{52}C_1 + 0.95^{52}\right) \\= 0.74$

Comments

Shouldn't it be
1 - (0.95^51 X 0.05 X 52 C 51 + 0.95^52 ) ?

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Isnt both the same?

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Yes They are! Actually Notation caused the confusion. Everything clear

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Also shouldn’t there be the case when 50 passengers show up?

it should be (case when 50 passengers show up) + (case when 51 passengers show up) + (case when 52 passengers show up)         ?

Answer 2

This is a question from Sheldon Ross,

Let’s start,

P(A person is present) = 0.95

P(A person is absent)  = 0.05, these are given in the question.

We must find the probability that everyone who comes will get their seats, but we know that there are only 50 seats and in the worst case 52 0r 51 people will come.

Instead of finding the probability of success, let’s find the probability of failure and subtract it from 1.

p(all 52 will come) =$0.95^{52}$ = 0.069

p(exactly 51 will come) = $\binom{52}{1}$$(0.05)^{1}$$(0.95)^{51}$ = 0.190

->1- (0.069 + 0.1900)

->0.74

 

Comments

Also shouldn’t there be the case when 50 passengers show up?

it should be (case when 50 passengers show up) + (case when 51 passengers show up) + (case when 52 passengers show up)