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Thread: Independent probabilities question

  1. #1
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    Independent probabilities question

    Dear All,

    I am new to probability theory and it would be a great help if you can help me answering this question.


    There are N balls. There is a circle in the space. The probability of each ball to fall in the circle is p. What is the probability that there are *at least* k out of N balls that fall in the circle? I came up with the probability of at least one out of N balls falling in the circle and that is ( 1 - (1 - p)^N). Can someone please guide me how to do for k.


    Thanks ,
    Cheema
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  2. #2
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    Hello cheema

    Welcome to Math Help Forum!
    Quote Originally Posted by cheema View Post
    Dear All,

    I am new to probability theory and it would be a great help if you can help me answering this question.


    There are N balls. There is a circle in the space. The probability of each ball to fall in the circle is p. What is the probability that there are *at least* k out of N balls that fall in the circle? I came up with the probability of at least one out of N balls falling in the circle and that is ( 1 - (1 - p)^N). Can someone please guide me how to do for k.


    Thanks ,
    Cheema
    Looking at the title of your post, I'm assuming that the ways in which the balls fall are independent of each other, so this is an example of a Binomial Distribution.

    The probability of each trial resulting in a success is $\displaystyle p$, where a 'trial' is a ball falling and a 'success' is the ball falling into the circle. The number of trials is $\displaystyle N$, and we want the probability that the number of successes is greater than or equal to $\displaystyle k$.

    The formula that gives the probability of getting exactly $\displaystyle r$ successes in N trials is:
    $\displaystyle \binom Nrp^r(1-p)^{N-r}$
    and we shall need the sum of all the terms like this, for values of $\displaystyle r$ from $\displaystyle k$ to $\displaystyle N$; in other words:
    $\displaystyle \sum_{r=k}^N\binom Nrp^r(1-p)^{N-r}$
    And that's it.

    Grandad
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  3. #3
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    That helped a lot. Thanks for answering in so clearly
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