[Sasa Luke]

Hello brains!please help me to solve this probability problem.i'm kinda hate this but since i must re-sit this paper,I must learn.here's the question.


Q) A certain airline company,having observed that 5% of people making reservations on flight do not show up for the flight,sells 100 seats on the plane that has only 98 seats.What is the probability that there will be a seat available for every person who shows up for the flight?


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[dantheman1261]

Edit: The solution below is certainly what is wanted

[Sasa Luke]

(Original post by dantheman1261)
Any more information given? If not then this isn't possible.

That's all given.This is past year question.How terrible it is!btw,could u help me with this one:


Q. If X1 and X2 are independent normal variables with mean 23 and 4 ,and variance 3 and 1 respectively,what distribution does (4X1-4X2) have?

I guess this one will be using normal distribution.pleaseeee help!thanks anyway!


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[dantheman1261]

(Original post by Sasa Luke)
That's all given.This is past year question.How terrible it is!btw,could u help me with this one:


Q. If X1 and X2 are independent normal variables with mean 23 and 4 ,and variance 3 and 1 respectively,what distribution does (4X1-4X2) have?

I guess this one will be using normal distribution.pleaseeee help!thanks anyway!

Unfortunately you just have to know that:

If is a normal random variable with mean and variance , then is normal with mean and variance .

If you add together two normally distributed random variables, then you get a normal random variable, and you just add the means to get the new mean, and add the variances to get the new variance.

[msmith2512]

(Original post by Sasa Luke)
Hello brains!please help me to solve this probability problem.i'm kinda hate this but since i must re-sit this paper,I must learn.here's the question.


Q) A certain airline company,having observed that 5% of people making reservations on flight do not show up for the flight,sells 100 seats on the plane that has only 98 seats.What is the probability that there will be a seat available for every person who shows up for the flight?


This was posted from The Student Room's iPhone/iPad App

There will be a seat for everyone if 98 or less people turn up. i.e 1 - probability that 99 or 100 people turn up.

Can you calculate prob(exactly 100 people turn up) and prob(exactly 99 people turn up)?

[dantheman1261]

(Original post by msmith2512)
There will be a seat for everyone if 98 or less people turn up. i.e 1 - probability that 99 or 100 people turn up.

Can you calculate prob(exactly 100 people turn up) and prob(exactly 99 people turn up)?

Not without some distribution giving these probabilities. There's no modelling information, or even an assumption of independence.

[notnek]

(Original post by dantheman1261)
Not without some distribution giving these probabilities. There's no modelling information, or even an assumption of independence.

You can assume independence and model it yourself.

Let X be the number of passengers who do not show up.

X~B(100,0.05)

It's been a while since I did this kind of thing but I remember that independence and modelling information were not always given in questions.

[msmith2512]

(Original post by dantheman1261)
Not without some distribution giving these probabilities. There's no modelling information, or even an assumption of independence.

Sorry you can calculate this. The question states that a person doesn't turn up with probability 5%. Given the brief nature of the question I think you have to assume that these are independent (in reality there will be groups of people travelling together).

So this is a binomial distribution and can be calculated.

[Nice.Guy]

It sounds like Poisson Distribution? OP are you sure it doesn't say which distribution this is, further up in the question?
Anyway, I think it's poisson...

assume X is talking about the number of people who DON'T turn up.
average number of people (mean) who don't turn up = np = 100*0.05 = 5
using the equation e^-lamda * lamda^r
--------------------- , you get P(X=0) = 6.74x10^-3 and P(X=1) = 0.0337
r!
take them both away from 1,
got an answer of about 0.96, sounds about right??

[msmith2512]

(Original post by Nice.Guy)
It sounds like Poisson Distribution? OP are you sure it doesn't say which distribution this is, further up in the question?
Anyway, I think it's poisson...

assume X is talking about the number of people who DON'T turn up.
average number of people (mean) who don't turn up = np = 100*0.05 = 5
using the equation e^-lamda * lamda^r
--------------------- , you get P(X=0) = 6.74x10^-3 and P(X=1) = 0.0337
r!
take them both away from 1,
got an answer of about 0.96, sounds about right??

See previous two posts this is a binomial distribution. As will be the case when there are two outcomes to an event. i.e turn up (prob = p) and don't turn up (prob = 1-p).

Since n is large and (1-p) small this will approximate to poisson and this answer is close to being correct.

The exact answer is 1-[(0.95)^100 + 100*(0.95)^99*0.05].

[Nice.Guy]

(Original post by msmith2512)
See previous two posts this is a binomial distribution. As will be the case when there are two outcomes to an event. i.e turn up (prob = p) and don't turn up (prob = 1-p).

Since n is large and (1-p) small this will approximate to poisson and this answer is close to being correct.

The exact answer is 1-[(0.95)^100 + 100*(0.95)^99*0.05].

Right, thanks, so you can't just assume Poisson if they don't specify... so was my answer an alternative method, or was it just coincidence it was close then??

Also, could you briefly explain the second bit in the bracket please? The 100*(0.95)^99*0.05 and is this AS or A2??

[msmith2512]

As n is large and (1-p) is small your solution was a good approx. however I am not sure you quite understood why.

My reasoning was prob(100 people turn up) = (0.95)^100.

Prob(99 people turn up) = 100*(0.95)^99*0.05. Being the number of ways 1 person doesn't turn up I.e. 100 (# ways of choosing 1 person from 100) multiplied by the probability which is (0.95)^99*(0.05). 99 people turn up with prob 0.95; and 1 doesn't with prob 0.05.

Prob(98 or fewer) = 1 - prob(exactly 100)-prob(exactly 99).

This is one of the stats modules so probably A2.

[ghostwalker]

Small point:

If you subtract two independent normal distributions, then you subtract the means, but add the variances.