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I am just starting to study probability theory, this question may sound naive but I would like to have some qualitative explanation for the below problem. There are 7 balls in a bag. Two balls are drawn at random. What is the probability of each ball to be second ball to be drawn. I guess it is 1/7 because any ball could be a second ball. But I am also confused on why it should not be 1/6, since it will be just 6 balls when second ball is drawn. Can someone please help me understand this? How can I derive this using probability formula?

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  • $\begingroup$ It may be that the probability stays unchanged ($1/7$) if having drawn the first ball doesn't provide you with any additional information other than the experiment is in progress, as opposed to the situation where you actually get to see the first draw, which would turn your problem into a conditional probability of the type $\text{p}(2^\text{nd}=\text{ball i}\,|\,1^{st}=\text{ball j})$. $\endgroup$ – Antoni Parellada Mar 26 '16 at 18:50
  • $\begingroup$ Whatever your argument may be, it will apply with equal force to every ball in the bag. If you conclude that each ball has $1/6$ chance of being drawn second, then the axioms of probability will compel you to assign $7/6$ chance to some ball being drawn second. That's a problem. $\endgroup$ – whuber Mar 26 '16 at 23:26
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Here's how to solve it, with some information left out since this probably falls under self-study.

\begin{align} P(\text{Ball }i \text{ is second}) &= P(\text{Ball }i \text{ is second} \mid \text{Ball }i\text{ is not first})P(\text{Ball }i\text{ is not first}) \\ &= \frac{x}{y}\frac{6}{7}. \end{align}

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