PDF's: Binomial Formula or Pascal's Formula

[whitejac]

Homework Statement

50 students live in a dormitory. The parking lot has the capacity for 30 cars. If each student has a car with probability 12 (independently from other students), what is the probability that there won't be enough parking spaces for all the cars?

Homework Equations

P(A) = P(B)P(C)
Binomial: C(n k) P^k(1-p)^n-k
Pascal's: C(k-1 m-1) P^m(1-p)^k-m

3. The Attempt at a Solution
So the difference is in the coefficient, clearly, and I'm wondering which one to use in this circumstance. The first time i did it, I chose the Binomial because it looks like a classic "What's the probability of this many trials having successes"? type of thing, but what if you defined the problem as "what's the probability of taking 49 trials, and getting 29 successes (kids having cars for the parking lot) and then on the 50th receiving a pass. I chose 29 successes because we are tasked to find the chance of having enough spaces, not more. The difference may be negligible when you do the factorials. After all, the middle numbers will all be the same, just the ones on the ends will be slightly altered, but I really do not see the difference in use between these two formulas. It sounds like two mathematical perspectives of the same thing.

 

[HallsofIvy]

"Probability 12"? Did you mean 0.12? In any case, yes this is a basic binomial probability problem but your numbers are a little off. The problem is not about "29 successes in 49 trials" but rather the probability of less than or equal to 30 successes in 50 trials. I think 50 trials is large enough that you can use the normal distribution approximation.

 

[Ray Vickson]

Homework Statement

50 students live in a dormitory. The parking lot has the capacity for 30 cars. If each student has a car with probability 12 (independently from other students), what is the probability that there won't be enough parking spaces for all the cars?

Homework Equations

P(A) = P(B)P(C)
Binomial: C(n k) P^k(1-p)^n-k
Pascal's: C(k-1 m-1) P^m(1-p)^k-m

3. The Attempt at a Solution
So the difference is in the coefficient, clearly, and I'm wondering which one to use in this circumstance. The first time i did it, I chose the Binomial because it looks like a classic "What's the probability of this many trials having successes"? type of thing, but what if you defined the problem as "what's the probability of taking 49 trials, and getting 29 successes (kids having cars for the parking lot) and then on the 50th receiving a pass. I chose 29 successes because we are tasked to find the chance of having enough spaces, not more. The difference may be negligible when you do the factorials. After all, the middle numbers will all be the same, just the ones on the ends will be slightly altered, but I really do not see the difference in use between these two formulas. It sounds like two mathematical perspectives of the same thing.

No, they are not even remotely the same. The Pascal (usually called the "negative binomial" nowadays) is the pmf of the trial number on which a certain number of successes first occurs in a sequence of Bernoulli trials. So $C(49,29) (.12)^{29} (.88)^{20}$ is the probability that Student 49 is the first one to occupy the parking space 29. That is not the same as asking for the probability that the first 49 students occupy no more than 29 spaces; can you see why they are different? (In fact, though, even that latter probability is not really relevant to the problem.)

This problem needs the binomial distribution, and you want $P(X \leq 30)$ for $X$ a binomial random variable with parameters $n = 50, p = 0.12$ (assuming your "12" was meant to be 0.12).

 

[whitejac]

Ohhh, okay. I think I understand now. My book portrayed it as "if you flip a coin and you want to see how many flips you can make before you land a success." So, I wondered why I couldn't equally represent this problem like this: You want to know how many students don't have a car before someone does - that someone being the 30th person. if the person who began having a car was identified after we'd already established the parking spaces were sufficient then it wouldn't matter. I can see now how that would be a little different. We are limiting ourselves to just one possibility of the possible probabilities to distinguish 30 people getting cars. Binomial Formula was my first guess - where
n = 50
k = 30
For we wish to know the possibility of 30 people receiving cars and 20 not receiving cars.

Also, I apologize for the confusion about the twelve. That was a copy-and-pasting error. P = 1/2 = q but the actual values weren't really my confusion.

 

[whitejac]

(More formally) The way I view it is... if we're tasked to find the probability that the parking lot will be big enough for the 30 kids with cars, then we could say:
PX = probability of x kids getting cars
PX(x) = { C(50 x) p^x(1-p)^{50 - x} for x = 0,1,2,3,4,... 50
0 otherwise }

And, in this case, PX(x ≤ 30) which would be the sum of PX(x) for x = 1,2,3...30.

 

[Ray Vickson]

(More formally) The way I view it is... if we're tasked to find the probability that the parking lot will be big enough for the 30 kids with cars, then we could say:
PX = probability of x kids getting cars
PX(x) = { C(50 x) p^x(1-p)^{50 - x} for x = 0,1,2,3,4,... 50
0 otherwise }

And, in this case, PX(x ≤ 30) which would be the sum of PX(x) for x = 1,2,3...30.

Actually, you need to include the term for x = 0, but it will be small.