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# Need urgent check on Probability Question watch

1. An airline finds that two thirds of the tickets from London to the island of Martinados are purchased by British passport holders. Each British passenger has a 60% chance of needing to catch a conn.ecting flight from Martinados, whereas this probability is only 10% for non-British passengers.

(i) A passenger on the flight is selected at random. What is the probability that this
passenger is transferring to a connecting flight at Martinados?

Answer: (2/3 * 0.6) + (1/3 * 0.1) = 13/30

(ii) Of those who are transferring to a connecting flight, what proportion hold British passports?

P(British|Connecting Flight) = (2/3 * 0.6)/(13/30) = 12/13

(iii) Use a suitable approximation to calculate the probability that, on a plane which holds 400 passengers, there are at least 150 British passport holders who are transferring to a connecting flight.
.
This one i have forgotten. I have done the following though:

Theta(hat) = X/k = 150/400 = 0.375 = Proportion at least 150 B-passengers on plane.

now I assume it is ~N(0.375, [(0.375)(1-0.375)/400]) and also that the
P(British Passengers and Connecting Flight) = 0.4

so z = (0.375 - 0.4)/sqrt(variance above) = -1.03

so P(Proportion of British Passengers > 0.375) = 0.85

If you can't answer iii) due to knowledge limit, please try ii) as that should be approx sixth form level
2. (Original post by mathsRus)
An airline finds that two thirds of the tickets from London to the island of Martinados are purchased by British passport holders. Each British passenger has a 60% chance of needing to catch a conn.ecting flight from Martinados, whereas this probability is only 10% for non-British passengers.

(i) A passenger on the flight is selected at random. What is the probability that this
passenger is transferring to a connecting flight at Martinados?

Answer: (2/3 * 0.6) + (1/3 * 0.1) = 13/30

(ii) Of those who are transferring to a connecting flight, what proportion hold British passports?

P(British|Connecting Flight) = (2/3 * 0.6)/(13/30) = 12/13

(iii) Use a suitable approximation to calculate the probability that, on a plane which holds 400 passengers, there are at least 150 British passport holders who are transferring to a connecting flight.
.
This one i have forgotten. I have done the following though:

Theta(hat) = X/k = 150/400 = 0.375 = Proportion at least 150 B-passengers on plane.

now I assume it is ~N(0.375, [(0.375)(1-0.375)/400]) and also that the
P(British Passengers and Connecting Flight) = 0.4

so z = (0.375 - 0.4)/sqrt(variance above) = -1.03

so P(Proportion of British Passengers > 0.375) = 0.85

If you can't answer iii) due to knowledge limit, please try ii) as that should be approx sixth form level
1 and 2 are both correct. (Why do you think it's too big? The proportion of non-Brits with a connecting flight is obviously tiny, and there are more Brits anyway.)

Third one: your working is a bit obscure, and I'm really not sure why you've got roughly the right answer. Why can you assume that follows that normal distribution? (I'm really bad at stats, so it may be obvious.)

I'd phrase it as:

P(British and connecting) = 2/5
so we seek P(X >= 150) where X follows the binomial(400, 2/5) distribution.

Approximating to the normal, we require the probability that Y > 149.5, where Y follows N(400*2/5, 400*3/5*2/5) = N(160, 96).

In terms of the standard normal, that is the probability that Z > -10.5/sqrt(96) = -1.072, given that Z follows N(0, 1). That turns out to be 0.858.
3. (Original post by Smaug123)
1 and 2 are both correct. (Why do you think it's too big? The proportion of non-Brits with a connecting flight is obviously tiny, and there are more Brits anyway.)
Thank you very much!

For the last one, I did think it was related to ~Bin(n,p) but did not know which 'n' to use.
4. (Original post by mathsRus)
Thank you very much!

For the last one, I did think it was related to ~Bin(n,p) but did not know which 'n' to use.
You're selecting passengers out of the 400 on a plane. It couldn't really be anything other than 400.
5. (Original post by Smaug123)
You're selecting passengers out of the 400 on a plane. It couldn't really be anything other than 400.
True, it does make sense. I assumed ~N(0,1) because sample was big enough.

Also, I have exam tomorrow, and my lecturer hasn't uploaded mark scheme so I have to post on here.

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