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Math Help - Probability confusion.

  1. #1
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    Probability confusion.

    There are two types of car in a company car fleet. Car type A (say there are 10 cars) and car type B (say there are also 10 cars). If a car from car type A breaks down, does this make it less likely that a car in car type B would break down in the same week? Another way of putting it is ... say there are a group of cars which we haven't differentiated (which are generally reliable) and 3 have broken down in one week (very rare), I assume it would be quite unlikely that another one would break down in the same week (i.e. the probability decreases each time - am I correct in this assumption?). But if those 3 cars are all car type A, and the fourth car we are observing is car type B, would this car be any more likely to break down or not, considering the back 'luck' of the past week, and considering we have now differentiated it to another group of cars?
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  2. #2
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    Assuming the cars are all independent of one another, i.e. any given car has probability p of breaking down, regardless of what happens to the other cars...

    Then two cars breaking down is two independent events, so to find the probability of it happening, you multiply the probabilities; p x p = p^2 ... less than the probability of one car breaking down

    However, (I think this is where you're getting confused) if a car has ALREADY broken down, the probability that another will break down is still p, because they are independent.

    If they aren't independent you need to know how they are related in order to answer questions about them
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  3. #3
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    Quote Originally Posted by newbie2008 View Post
    There are two types of car in a company car fleet. Car type A (say there are 10 cars) and car type B (say there are also 10 cars). If a car from car type A breaks down, does this make it less likely that a car in car type B would break down in the same week? Another way of putting it is ... say there are a group of cars which we haven't differentiated (which are generally reliable) and 3 have broken down in one week (very rare), I assume it would be quite unlikely that another one would break down in the same week (i.e. the probability decreases each time - am I correct in this assumption?). But if those 3 cars are all car type A, and the fourth car we are observing is car type B, would this car be any more likely to break down or not, considering the back 'luck' of the past week, and considering we have now differentiated it to another group of cars?
    newbie2008,

    As others have already pointed out, if the cars are independent then one's breaking down has no effect on the probability that another will break down.

    However, independence of events is simply a simplifying mathematical assumption, and may or may not happen in the real world. There are many real world factors that could falsify this assumption. For example, all cars are subject to the same weather and general road conditions. If it's been hot lately then all the cars are probably running their air conditioners, putting more stress on the engine and making a breakdown (of the air conditioner or something else) more likely. If the weather has been icy then there are more potholes in the streets, increasing the likelihood of damage to all cars. If one car breaks down then there may be increased use of the other cars, putting more wear and tear on them and making their breakdown more likely.

    The bottom line is this: Independence may or may not be present in real world events. Mathematics alone cannot tell you if events are independent, because mathematics knows nothing of the real world.
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