MHB Probability Question: At Least One Customer Unable To Claim Car

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In a car rental scenario, 20% of reservations are not claimed, with 25 reservations made for 22 available cars. The probability that all cars are claimed is calculated to be approximately 0.2339, while the probability that 23 or more reservations show up is about 0.098. The discussion centers on understanding how to derive the relationship between these probabilities, particularly how the probability of one or more customers being unable to claim their car relates to the event of all cars being claimed. The conclusion is reached that the probability of one or more reservations not being able to claim a car, given that all cars are claimed, is found by dividing the two probabilities. This results in a clearer understanding of the conditional probability involved in the situation.
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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) $.
 
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tmt said:
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"?
 
Jameson said:
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.
 
Last edited:
tmt said:
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:
$$
P(\text{one-or-more-reservations-not-able-to-claim} \land \text{all-cars-claimed}) \\
= P(Claims > 22 \land Claims \ge 22) \\
= P(Claims > 22) \\
= \binom{25}{23} 0.8^{23}0.2^2 + \binom{25}{24} 0.8^{24}0.2^1 + \binom{25}{25} 0.8^{25}0.2^0
$$
?
 
I like Serena said:
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)
 
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