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    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

    Answer seems too big.

    (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
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    (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

    Answer seems too big.

    (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 \hat{\theta} 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.
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    (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.
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    (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.
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    (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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