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    Exclamation Probability Question

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    1.Can anyone explain how to solve this ???

    A fair die (sides having numbers 1-6) is to be rolled until the first number greater than 4 appear.What is the probability that first number greater than 4 will appear on 3rd or 4th roll.


    Ans : 20/81

    2. Container contains 10 disks each numbered 1 thr 10 inclusive.If 4 disks are to be selected one aftr the other with each disk selected at random without replacement.What is the probability that the range of numbers on disk selected is 7 ?

    Ans : 3/14
    Last edited by The Rain; 09-14-2012 at 01:54 AM.

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    Within my grasp! thepillow's Avatar
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    1. On any given roll the odds of getting a number greater than 4 is 2/6 = 1/3. (Only 5 and 6 are greater than 4, so this is 2 out of the 6 numbers.)

    This means that the probability of getting a number that is 4 or less is 2/3. We'll use these numbers in a little bit.

    We want to consider 2 cases:

    Case 1: We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 and our second roll is less than or equal to 4 and our third roll is greater than 4.

    We can write this as: P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (1/3)

    Case 2: We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 and our second roll is less than or equal to 4 and our third roll is less than or equal to 4 and our fourth roll is greater than 4.

    We can write this as: P(4 or less) x P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (2/3) x (1/3)

    We want to know about either case happening so we can add up these probabilities: (2/3)(2/3)(1/3) + (2/3)(2/3)(2/3)(1/3) = 4/27 + 8/81 = 12/81 + 8/81 = 20/81


    2. In order for there to be a range of 7 we need the largest number minus the smallest number to equal 7. From the numbers between 1 and 10, we can make 3 different pairs that satisfy this:

    8 and 1; 9 and 2; 10 and 3

    Let's take 8 and 1 as an example: along with our 8 and 1, we need 2 other discs out of the 6 numbers between 8 and 1 (we can't get a 9 or a 10 because that would change the range). So there are 6c2 = 15 different pairs we could choose for our other 2 discs.

    This works out the exact same way for 9 and 2, as well as 10 and 3. So in each of the 3 cases there are 15 combinations for a total of 45.

    The number of total groups of 4 that we can get from the 10 discs is 10c4 = 210.

    So 45 of the 210 groups meet our requirements. 45/210 can be reduced to 3/14.


    Hope that helps! Let me know if I need to explain something better.

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    Quote Originally Posted by thepillow View Post
    1. On any given roll the odds of getting a number greater than 4 is 2/6 = 1/3. (Only 5 and 6 are greater than 4, so this is 2 out of the 6 numbers.)

    This means that the probability of getting a number that is 4 or less is 2/3. We'll use these numbers in a little bit.

    We want to consider 2 cases:

    Case 1: We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 and our second roll is less than or equal to 4 and our third roll is greater than 4.

    We can write this as: P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (1/3)

    Case 2: We get a number greater than 4 for the first time on the 3rd roll = Our first roll is less than or equal to 4 and our second roll is less than or equal to 4 and our third roll is less than or equal to 4 and our fourth roll is greater than 4.

    We can write this as: P(4 or less) x P(4 or less) x P(4 or less) x P(greater than 4) = (2/3) x (2/3) x (2/3) x (1/3)

    We want to know about either case happening so we can add up these probabilities: (2/3)(2/3)(1/3) + (2/3)(2/3)(2/3)(1/3) = 4/27 + 8/81 = 12/81 + 8/81 = 20/81


    2. In order for there to be a range of 7 we need the largest number minus the smallest number to equal 7. From the numbers between 1 and 10, we can make 3 different pairs that satisfy this:

    8 and 1; 9 and 2; 10 and 3

    Let's take 8 and 1 as an example: along with our 8 and 1, we need 2 other discs out of the 6 numbers between 8 and 1 (we can't get a 9 or a 10 because that would change the range). So there are 6c2 = 15 different pairs we could choose for our other 2 discs.

    This works out the exact same way for 9 and 2, as well as 10 and 3. So in each of the 3 cases there are 15 combinations for a total of 45.

    The number of total groups of 4 that we can get from the 10 discs is 10c4 = 210.

    So 45 of the 210 groups meet our requirements. 45/210 can be reduced to 3/14.


    Hope that helps! Let me know if I need to explain something better.

    Understood !
    Thanks for explaining nicely.
    Any tips how to solve such problems in short time ?

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    Within my grasp! thepillow's Avatar
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    Quote Originally Posted by The Rain View Post
    Any tips how to solve such problems in short time ?
    That's the million dollar question!

    Well, I think the first step should always be to "translate" the question into more intuitive and colloquial language. This might seem trivial, but unless it is immediately apparent to you what the question is asking, this little trick can make a huge difference. If you can express the same question in different words, then you have almost certainly truly understood what it is asking.

    I'll use one of your original questions as an example:

    "A fair die (sides having numbers 1-6) is to be rolled until the first number greater than 4 appear.What is the probability that first number greater than 4 will appear on 3rd or 4th roll?"

    The language here isn't too convoluted, but we can take the question apart and rephrase it so it is more obvious what's going on. The first part - "a fair die (sides having numbers 1-6) is to be rolled" - is easily understood, but let's just change it to "we roll a die". We can assume it's fair unless told otherwise. Then we have "until a number greater than 4 appears". This can be restated as "until getting a 5 or a 6". The next part - "the first number greater than 4 will appear on the 3rd or 4th roll" - can be "translated" as "this takes us 3 or 4 rolls".

    Now the whole question is: We roll a die until getting a 5 or a 6. What is the probability that this takes 3 or 4 rolls?

    Again, the original question was: "A fair die (sides having numbers 1-6) is to be rolled until the first number greater than 4 appear.What is the probability that first number greater than 4 will appear on 3rd or 4th roll?"

    In my opinion our new version is much easier to deal with. But they both say the same thing!

    It might seem like this type of translation is a slow process and that it will waste time on the test. But with a bit of practice this can become second nature and can end up saving you time.

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