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Two balls are chosen at random from a box containing 12 balls, numbered 1, 2, ..., 12. Let X be the larger of the two numbers obtained. Compute the PMF of X, if the drawing is done a.) with replacement b.) without replacement

I'm having trouble with finding an easier way to solve the second part of this problem. I understand that the total number of outcomes is 12*12, this is the denominator, but the numertaor, favorable outcomes, is giving me trouble. I know the answer is 2x-1 but I figured this out by seeing a pattern for each possible outcome. Is there an easier way to deduce this answer? How can I get that there is 2x-1 without haveing to write out every single outcome and looking for a pattern? Please don't just say there are 2x-1 choices. I want to understand the thought process. In otherwords how did you arrive at this.

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Let $X=\max\{X_1, X_2\}$ where $X_1, X_2$ are the results of the two draws.

The outcomes for the event are those where "one draw equals $x$ the other less, or both draws equal $x$".

When drawing with replacement, the results for each draw is independent.

$$\begin{align}\newcommand{\P}{\operatorname {\sf P}}\P(\max\{X_1,X_2\} = x) & =\mathsf P(X_1=x, X_2\leq x)+\P(X_1<x,X_2=x)\\[2ex] & =\mathsf P(X_1=x)\P(X_2\leq x)+\P(X_1<x)\P(X_2=x) \\[1ex] & = \qquad\frac{1}{12}\quad\cdotp\quad\frac{x}{12}\qquad+\qquad\frac{x-1}{12}\quad\cdotp\quad\frac{1}{12}\\[1ex] & = \frac{2x-1}{144}\end{align}$$

When drawing without replacement, the results are dependent, and in particular, the tie cannot happen.

$$\begin{align}\newcommand{\P}{\operatorname {\sf P}}\P(\max\{X_1,X_2\} = x) & =\mathsf P(X_1=x, X_2< x)+\P(X_1<x,X_2=x)\\[2ex] & =\mathsf P(X_1=x)\P(X_2< x\mid X_1=x)+\P(X_1<x\mid X_2=x)\P(X_2=x) \\[1ex] & = \quad\qquad\frac{1}{12}\quad\cdotp\quad\frac{x-1}{11}\qquad+\qquad\frac{x-1}{11}\quad\cdotp\quad\frac{1}{12}\\[1ex] & = \frac{2(x-1)}{132}\end{align}$$

That is all.

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  • $\begingroup$ @Allan The keyword is: Independence. When you replace the ball, then the result of the second draw does not depend on the result of the first draw; so there is no conditioning. $\endgroup$ – Graham Kemp Feb 19 '17 at 16:39
  • $\begingroup$ Yeah I came to that realization right after I posted the question. Thanks @Graham Kemp this really cleared up a lot of concepts for me. $\endgroup$ – Allan Feb 19 '17 at 16:50

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