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Math Help - probability of a selection

  1. #1
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    probability of a selection

    Hi,

    A bag has 40 tickets number from 1 to 40. 4 tickets namely t1,t2,t3,t4 have been taken and arranged as t1<t2<t3<t4.

    Then what is the probability of having t3 as 20.

    What i have tried is, t1 can range from 1 to 19. so the t1's probability is 19/40.

    t2 can be remaining 18 numbers which are less than 20, and total tickets are 39 now. so 18/39.

    And t3 is 1/38 and t4 can be any one of the remaining 37 tickets so 1/37.

    The probability of the interested event is 19/40 * 18/39 * 1/38 * 20/37.

    But, this is not matching with the answer.

    Please correct if my approach is wrong.

    thanks.
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  2. #2
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    You under counted the possibilities.
    Any selection of four tickets can be arranged is ascending order.
    There are \binom{40}{4} such selections.
    How of those contain the ticket numbered 20, two ticket numbered less than 20 and one numbered greater than 20?
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  3. #3
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    Thanks.

    the answer would be, (19C2 * 20C1) / 40C4.


    But, how should one interpret my approach. I don't understand what i have tried there.

    I appreciate, if you could explain.
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  4. #4
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    Quote Originally Posted by kumaran5555 View Post
    But, how should one interpret my approach. I don't understand what i have tried there.
    I don't understand that approach either.
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  5. #5
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    Hello, kumaran5555!

    The answer would be: . \displaystyle \frac{(_{19}C_2)(_{20}C_1)}{_{40}C_4}

    But, how should one interpret my approach.
    I don't understand what i have tried there.

    Now, I don't understand!
    You wrote that answer . . . but don't know how or why?


    There are _{40}C_4 possible outcomes.
    . . That is the denominator.


    Now consider the numerator. .(Plato already outlined it.)

    We want "20" . . . There is one (1) way.

    We want two numbers to be less than 20.
    . . There are 19 numbers less than 20; select 2 of them.
    .There are _{19}C_2 ways.

    We want one number to be greater than 20.
    . . There are 20 numbers greater than 20; select 1 of them.
    . . There are _{20}C_1 ways.

    The numerator is: . 1\cdot(_{19}C_2)\cdot(_{20}C_1)


    Got it?

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  6. #6
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    Thanks Soroban.

    But I want to know the interpretation of my first approach, which I have showed in the first post of this thread.

    I know, I have really messed up there. But to make myself clear, i need to know what i have done there.

    So please if you could check.
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  7. #7
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    As I have told you, I don’t know what you were thinking in that first post.
    It really makes no sense.
    Although this may help you: Order makes no difference in this question.
    We make a selection of four, one of which is 20, two less than 20 and one more.
    You may have been thinking order matters?
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