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In a vase a 10 balls: 4 red and 6 white. They are taken out one by one without replacement.

Let X be the stochastic variable that denotes how often we need to draw balls before we draw a white ball.

Calculate the probability function of X.

To answer this question I wanted to use the geometric probability function, $q^{k-1}p$, but this would only work if the 'experiments' (drawing a red ball is a failed experiment, drawing a white ball a successful experiment) are independent, which, in this case, it isn't since we are drawing without putting the balls back.

How can I approach this to find an answer?

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1 Answer 1

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Clearly, $X$ is between 1 and 5. The probability that $X$ equals 1 is the probability of drawing a white ball on the first try, thus $\mathbb{P}(X=1)=6/10=.6$.

Similarly, the probability that $X$ equals 2 is the probability of drawing first a red ball and then a white ball, thus $\mathbb{P}(X=2)=4/10\times 6/9$.

The probability that $X$ equals 3 is the probability of drawing first two red balls and then a white ball, that is $\mathbb{P}(X=3)=4/10\times 3/9\times 6/8$.

Can you complete the exercise?

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Yes, I think I can, thank you! Is this some variant of the geometric probability function (since for a geometric probability, the 'experiments' need to be independent), or something else completely? –  Garth Marenghi Aug 21 '13 at 17:51
    
It is not a variant of the geometric distribution. It is related in that it describes a similar situation (time of first occurrence of some event), but the fact that the probability of this event changes over time (because we don't replace balls) makes it quite different. –  Eckhard Aug 21 '13 at 20:52

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