Probability Question: At Least One Customer Unable To Claim Car

[tmt1]

So, at a car rental company, 20% of car reservations are not claimed. There is a total of 22 cars and the manager takes 25 reservations a day.

If all cars are claimed for a day, what is the probability that one or more customer who had reservations were unable to claim their car?

I need to find:

$P(one-or-more-reservations-not-able-to-claim | all-cars-claimed) = \frac{P(one-or-more-reservations-not-able-to-claim \land all-cars-claimed)} {P(all-cars-claimed)}$To get the probability that all cars are claimed, I take ${{25}\choose{22}} {0.8 ^{22} * 0.2^ 3}$ + ${{25}\choose{23}} {0.8 ^{23} * 0.2^ 2}$ + ${{25}\choose{25}} {0.8 ^{25} * 0.2^ 0}$ which is $P(all-cars-claimed) = 0.2339$.

The probability that 23 or more of reservations show up is $0.0982252228436887$

Based on the solution, I can infer that $0.0982252228436887 = P(one-or-more-reservations-not-able-to-claim \land all-cars-claimed) $. Thus, the answer is $0.0982252228436887 / 0.2339$.

But I'm not sure how to reason for myself that $0.0982252228436887 = P(one-or-more-reservations-not-able-to-claim \land all-cars-claimed) $.

 

[Jameson]

So, at a car rental company, 20% of car reservations are not claimed. There is a total of 22 cars and the manager takes 25 reservations a day.

If all cars are claimed for a day, what is the probability that one or more customer who had reservations were unable to claim their car?

Hi tmt, (Wave)

For the quoted part in bold, should this read "If all cars are reserved for the day..."? I think there are two events here, reserving and claiming.

How did you get "The probability that 23 or more of reservations show up is 0.0982252228436887"?

 

[tmt1]

Hi tmt, (Wave)

For the quoted part in bold, should this read "If all cars are reserved for the day..."? I think there are two events here, reserving and claiming.

How did you get "The probability that 23 or more of reservations show up is 0.0982252228436887"?

No -- there are 25 reservations every day. And there is a 0.8 probability that a reservation is claimed.

Therefore, each day it is expected that 20 cars will be claimed (25 * 0.8).

The probability that 23 or more of reservations are claimed ${{25}\choose{23}} {0.8 ^{23} * 0.2^ 2}$ + ${{25}\choose{24}} {0.8 ^{24} * 0.2^ 1}$ + ${{25}\choose{25}} {0.8 ^{25} * 0.2^ 0}$ which is $P(all-cars-claimed) = 0.098 $

Since there are only 22 cards, that means there is a 0.098 probability that 1 or more people that made a reservation do not actually get to claim their car.

To get the probability that all cars are claimed is the same as 22 or more reservations are claimed (if 22 reservations are claimed, then all available cars are claimed, thus if > 22 reservations are claimed, also all available cars are claimed).

This is ${{25}\choose{22}} {0.8 ^{22} * 0.2^ 3}$ + ${{25}\choose{23}} {0.8 ^{23} * 0.2^ 2}$ + ${{25}\choose{24}} {0.8 ^{24} * 0.2^ 1}$ + ${{25}\choose{25}} {0.8 ^{25} * 0.2^ 0}$ which is $P(all-cars-claimed) = 0.2339$.

Now, the question is if all cars are claimed for the day, what is the probability that 1 or more reservations are unable to claim their car?

Let call P(A) the probability that all cars are claimed for the day. Then $P(A) = 0.23$ from the work above.

Lets call P(B) the probability that 23 or more reservations make their claim. Then $P(B) = 0.098$.

So how do we answer this question?

The solution says 0.098 / 0.23 but I'm not sure how to intuit this.

 

[I like Serena]

But I'm not sure how to reason for myself that $0.0982252228436887 = P(one-or-more-reservations-not-able-to-claim \land all-cars-claimed) $.

Don't we have:
$$
P(\text{one-or-more-reservations-not-able-to-claim} \land \text{all-cars-claimed}) \\
= P(Claims > 22 \land Claims \ge 22) \\
= P(Claims > 22) \\
= \binom{25}{23} 0.8^{23}0.2^2 + \binom{25}{24} 0.8^{24}0.2^1 + \binom{25}{25} 0.8^{25}0.2^0
$$
?

 

[tmt1]

Don't we have:
$$
P(\text{one-or-more-reservations-not-able-to-claim} \land \text{all-cars-claimed}) \\
= P(Claims > 22 \land Claims \ge 22) \\
= P(Claims > 22) \\
= \binom{25}{23} 0.8^{23}0.2^2 + \binom{25}{24} 0.8^{24}0.2^1 + \binom{25}{25} 0.8^{25}0.2^0
$$
?

Oh yeah, finally clicked.

Simple a matter of P(A | B) = P(A ^ B) / P(B)