Hey there! Sign in to join this conversationNew here? Join for free

Computing probability - given some conditions Watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    I am doing probability and I am stuck on this question.

    A torch will last at least 500 hours from new with probability 0.6. If it laststhis long, it survives the next 500 hours with probability 0.4, and if it lasts1000 hours, the probability that it will fail within the next 500 hours is 0.9.What is the probability that the torch will last at least 1500 hours?

    I am kinda bad at probability but here is what I thought about it so far!
    The question gave some conditions and there are some formulas that involve conditions:
    - P(E|F) = P(EF)/P(F)
    - the multiplication rule: P(E1E2E3...En) = P(E1)P(E2|E1)P(E3|E1E2)...P(En|E1....En-1)
    - the total law of probability - i don't think its applicable here?

    Let A = event that torch lasts at least 1500 hours and I want to find P(A).
    Well from the question, we know that the probability of a torch failing within the next 500 hours is 0.9 if it lasts 1000 hours... so
    B = event that torch fails inbetween 1000 hours and 1500 hours
    C = event that torch lasted at 1000 hours
    P(B|C) = 0.9

    Can I just do 1 - 0.9 to get the probability that the torch lasted more than 1500 hours?

    I think my reasoning is somehow off... there are other numbers in the question I haven't used and I havent used any of the formulae...
    Offline

    13
    ReputationRep:
    (Original post by aspiring_doge)
    I am doing probability and I am stuck on this question.

    A torch will last at least 500 hours from new with probability 0.6. If it laststhis long, it survives the next 500 hours with probability 0.4, and if it lasts1000 hours, the probability that it will fail within the next 500 hours is 0.9.What is the probability that the torch will last at least 1500 hours?

    I am kinda bad at probability but here is what I thought about it so far!
    The question gave some conditions and there are some formulas that involve conditions:
    - P(E|F) = P(EF)/P(F)
    - the multiplication rule: P(E1E2E3...En) = P(E1)P(E2|E1)P(E3|E1E2)...P(En|E1....En-1)
    - the total law of probability - i don't think its applicable here?

    Let A = event that torch lasts at least 1500 hours and I want to find P(A).
    Well from the question, we know that the probability of a torch failing within the next 500 hours is 0.9 if it lasts 1000 hours... so
    B = event that torch fails inbetween 1000 hours and 1500 hours
    C = event that torch lasted at 1000 hours
    P(B|C) = 0.9

    Can I just do 1 - 0.9 to get the probability that the torch lasted more than 1500 hours?

    I think my reasoning is somehow off... there are other numbers in the question I haven't used and I havent used any of the formulae...
    Think about that definition of conditional probability, P(E|F) = P(E and F)/P(F). In this question, you have (for example) E = "Surviving for 1,500 hours" and F = "Surviving for 1,000 hours". Therefore the event (E and F) is the same as the event E. That is, for event E to happen, it must be the case that event F happens. Therefore that definition of conditional probability simplifies to P(E|F) = P(E)/P(F) and so you get P(E) = P(E|F)P(F).

    So,

    P(1,500) = P(1,500 | 1000) P(1000) = (0.1) x P(1000).

    Can you now apply the same reasoning to work out P(1,000)?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Gregorius)
    Think about that definition of conditional probability, P(E|F) = P(E and F)/P(F). In this question, you have (for example) E = "Surviving for 1,500 hours" and F = "Surviving for 1,000 hours". Therefore the event (E and F) is the same as the event E. That is, for event E to happen, it must be the case that event F happens. Therefore that definition of conditional probability simplifies to P(E|F) = P(E)/P(F) and so you get P(E) = P(E|F)P(F).

    So,

    P(1,500) = P(1,500 | 1000) P(1000) = (0.1) x P(1000).

    Can you now apply the same reasoning to work out P(1,000)?
    I am really confused when you said "the event (E and F) is the same as event E". Well this means that the torches lasted for 1,500 hours AND lasted for 1,000 hours? I still cant reason with why you said this is the same as E.
    [Side note: Does P(E|F) = P(E and F)/P(F) ALWAYS simplfy to P(E|F) = P(E)/P(F)?]

    Why did you rearrange to formula? I know that P(E|F) and P(E) aren't the same, its kind of hard for me to explain why I dont agree. Aren't we suppose to find the probability of E CONDITIONED on F?
    P(E|F) = P(E and F)/P(F)P(E|F) = 0.1/P(F)

    Now I find P(F):
    C = event that torch lasts 500 hours so P(F|C) = P(F and C)/P(C) and I guess P(F and C) somehow simplifies to P(F)
    [Question: I am confused here, I want to get P(F), but I would have to plug in 0.4 into P(F|C) which says "the probability that the torch last 1000 hours GIVEN that it lasts 500 hours", now what is the difference in saying he probability that the torch last 1000 hours AND that it lasts 500 hours" which says P(F and C)]

    Carrying on,
    0.4 = P(F)/0.6
    P(F) = 0.24

    So back to the first formula
    P(E|F) = 0.1/P(F)
    P(E|F) = 0.1/0.24
    P(E|F) = 0.875
    Offline

    3
    ReputationRep:
    Surely just:

    P(survive 500) x P(survive next 500) x P(survive next 500)

    0.6*0.4*0.1 = 0.024

    These can be treated as succesive independent outcomes, no?
    Offline

    13
    ReputationRep:
    (Original post by aspiring_doge)
    I am really confused when you said "the event (E and F) is the same as event E". Well this means that the torches lasted for 1,500 hours AND lasted for 1,000 hours?
    If a torch last for 1,500 hours, then it is automatically true that the torch lasted for 1,000 hours. So the compound event "the torch lasted 1,500 hours" AND "the torch lasted 1,000" hours is the same as the event "the torch lasted 1,500 hours".


    [Side note: Does P(E|F) = P(E and F)/P(F) ALWAYS simplfy to P(E|F) = P(E)/P(F)?]
    No. In general  P(E|F) = P(E \cap F)/P(F) and P(E \cap F) need not equal P(E)
    Offline

    13
    ReputationRep:
    (Original post by offhegoes)
    Surely just:

    P(survive 500) x P(survive next 500) x P(survive next 500)

    0.6*0.4*0.1 = 0.024

    These can be treated as succesive independent outcomes, no?
    Indeed. However, if an exercise is phrased in terms of conditional probabilities ("if it survives this long then the probability of it ..." ), it's a fair bet that demonstrating that you have a solid understanding of how to handle that conditioning (even in a simple example like this) will earn you a gold star.
    • Thread Starter
    Offline

    0
    ReputationRep:
    So I saw that offhegoes says the answer is 0.024.But I managed to get 0.21 by doing P(E) = P(E|F)P(F) => P(E) = 0.875 x 0.24
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Should Spain allow Catalonia to declare independence?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.