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Question :- From an urn containing $a$ white balls and $b$ black balls, a certain number of balls, $k$ is drawn and they are laid aside their colour unnoticed. Then one more ball is drawn. Find the probability that it is white.

My approach:-

Let $i$ be the number of white balls drawn

Therefore, the number of black balls drawn is $=$ $k-i$

Let E be when the ball is white

$F(i)$ $=$ "$i$ white balls are drawn"

$P(E) = \sum_{i=0}^kP(E\cap F_i)$

$P(E) = P(E \cap F_0)+P(E \cap F_1)+P(E \cap F_2)+...+P(E \cap F_k)$

$P(E) = \frac{a-0}{a+b-k}+\frac{a-1}{a+b-k}+\frac{a-2}{a+b-k}+...+\frac{a-k}{a+b-k}$

$P(E) = \frac{a+a-1+a-2+a-3+...+a-k}{a+b-k} $

$P(E) = \frac{a(k+1)-(1+2+3+4+...+k)}{a+b-k} $

$P(E) = \frac{a(k+1)-\frac{k(k+1)}{2}}{a+b-k} $

$P(E) = \frac{2a(k+1)-k(k+1)}{2(a+b-k)} = \frac{(k + 1)(2a - k)}{2(a + b - k)}$

The correct answer according to the book is $\frac{a}{a+b}$

What is the best way to work through this problem?

VOstro12
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  • With regards to your error, what you wrote was not $P(E\cap F_i)$ but rather was $P(E\mid F_i)$. You needed to multiply each by the probability $P(F_i)$ – JMoravitz Feb 02 '23 at 13:49
  • @JMoravitz Isn't P(E∩Fi) the same as P(Fi),P(E∣Fi)? Btw this question is not a duplicate one. https://math.stackexchange.com/questions/1287393/if-you-draw-two-cards-what-is-the-probability-that-the-second-card-is-a-queen is a completely different question. – VOstro12 Feb 02 '23 at 14:12
  • No, it is absolutely a duplicate. The only thing that changed was the flavor, but the point remains that initial draws which are not observed have zero influence on any draw(s) that follow and so the result you are after is going to be the same as though those initial unobserved draws did not happen at all. This is true regardless of whether we were talking about cards, whether we were talking about balls, or if we were talking about peanut butter sandwiches or whatever else... – JMoravitz Feb 02 '23 at 14:13
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    Yes, $P(E\cap F_i) = P(F_i)\cdot P(E\mid F_i)$. Your attempt was as though $P(E\cap F_i)=P(E\mid F_i)$ without the $P(F_i)$. You are missing the $P(F_i)$ in each of your terms in your calculation. Thanks to having forgotten to include those, your answer was off. – JMoravitz Feb 02 '23 at 14:14
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    Your answer should have been $\color{red}{\frac{\binom{a}{0}\binom{b}{k}}{\binom{a+b}{k}}}\frac{a-0}{a+b-k} + \color{red}{\frac{\binom{a}{1}\binom{b}{k-1}}{\binom{a+b}{k}}}\frac{a-1}{a+b-k}+\dots$ Those red terms were the $P(F_i)$ that you neglected to include. Yes, you could if you really insist go through all of this effort to write it all out and go through all the effort to algebraically simplify if you really wanted... This is largely a waste of time however, and the linked duplicate gives a way to avoid having to do any of that messy algebra and get straight to the answer by symmetry – JMoravitz Feb 02 '23 at 14:20
  • Each of the $a+b$ balls are just as likely to be the ball drawn that you look at... $a$ of which happen to be white. – JMoravitz Feb 02 '23 at 14:27

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