5
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We have a bag with $4$ white balls and $1$ black ball. We are drawing balls without replacement. Find expected value for the number of tries to draw the black ball from the bag.


Progress. The probability to draw a black ball from first trial is $1/5$. The problem is how to find the probability to draw black ball from $2$nd, $3$rd, $ \ldots, 5$th trial. When I know all this probabilities I can find expected value as $1\cdot(1/5) + 2 p_2 + \dots + 5 p_5$.

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  • $\begingroup$ Do you know the definition of expectation value? $\endgroup$ – Arthur Dec 18 '15 at 16:02
  • $\begingroup$ @Arthur Yes, from the wikipedia. The probability to draw a black ball from first tryal is 1/5. The problem is how to find the probability to draw black ball from 2nd, 3rd ... 5th tryal. When i know all this probabilities I can find expected value as 1*(1/5) + 2*p2 + ... + 5*p5 $\endgroup$ – Alexander Dec 18 '15 at 16:10
  • $\begingroup$ Exactly. Well, I invite you to answer your own question. $\endgroup$ – drhab Dec 18 '15 at 16:12
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    $\begingroup$ The probability is quite easy to figure out with this simple trick: Imagine you draw all the balls, one by one, and put them in a row, first to last. Then the black ball is equally likely to be in any if the five positions, which means that $p_1 = p_2 = p_3 =p_4 = p_5$. $\endgroup$ – Arthur Dec 18 '15 at 16:13
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    $\begingroup$ $p_2=\dfrac14\left(1-p_1\right)$ since you have four balls left, and $p_3=\dfrac13\left(1-p_1-p_2\right)$ and so on $\endgroup$ – Henry Dec 18 '15 at 16:17
5
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It is as if you will create a word with $4$ W's and $1$ B. For example $BWWWW$ or $WWWBW$ etc. How many such words can you create? Answer: $5$ and any such word is equally likely.

In other words: the probability that the black ball will be drawn at any place - not only the first - is equal to $1/5$. Not conditional probability, but probability. Do not get confused, that if you have drawn $4$ White balls then the probability of drawing the black ball in the fifth draw is $1$. This is the conditional probability. "A priori" it is equally likely that the black ball will be drawn at any given point from $1$ to $5$. So, $$E[X]=\frac{1}{5}\cdot 1+ \frac{1}{5}\cdot2+\ldots+\frac15\cdot 5=\frac15(1+2+3+4+5)=3 $$ (where $X$ denotes the number of trials).

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  • $\begingroup$ But when a black or a white ball has already been selected at the ith step, then shouldn't the probability change? $\endgroup$ – MathMan Oct 22 '19 at 18:09

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