Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

David
Lv 4
David asked in Science & MathematicsMathematics · 7 years ago

Maths Homework, Help!!?

An airline company accepts bookings for 20 executive class seats on a flight from London to Bergen. The probability that a customer, who has booked an executive class seat, will not turn up is 0.15 and may be assumed independent of the behaviour of other customers.

The airline also accepts bookings for 50 standard class seats on the same flight. The probability of a customer, who has booked an standard class seat, will not turn up is 0.08 and may be assumed independent of the behaviour of other customers.

The aeroplane has 19 executive class seats and 48 standard class seats. If 19 or fewer customers who have booked executive class seats turn up they will all be allocated executive class seats. If all 20 customers who have booked executive class seats turn up, one will be allocated a standard class seat. this will leave 47 seats available for the 50 customers who have booked standard class seats.

calculate the probability that there will be standard class seats available for all those who turn up having booked standard class seats.

---------------------------------------------------

I have tried various different attempts to answer this but my answers are all over the place. I would be very grateful if someone could show me how to do this question as i just can't seem to get it right.

1 Answer

Relevance
  • ?
    Lv 6
    7 years ago
    Favorite Answer

    P(enough stand available) =

    P(20 execs) P(47 or less stand) + P(19 or less execs) P(48 or less stand)

    P(20 execs) = (.85)^20 = (approx) .03876

    P(19 or less execs) = 1 - P(20 execs) = (approx) .96124

    P(50 stand) = (.92)^50 = (approx) .01547

    P(49 stand) = [50] (.92)^49(.08)^1 = (approx) .06725

    P(48 stand) = [(50)(49)/(2)] (.92)^48(.08)^2 = (approx) .14326

    P(48 or less stand) = 1 - P(50 stand) - P(49 stand) = 1 - .01547 - .06725 = .91728

    P(47 or less stand) = P(48 or less stand) - P(48 stand) = .91728 - .14326 = .77402

    P = (.03876)(.77402) + (.96124)(.91728) = .03000 + .88173 = .91173

    (answer approx)

Still have questions? Get your answers by asking now.