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(2 points) An urn contains 3 balls number 1 to 3. Two balls are drawn with replacement. Let X be the value of larger of the two numbers drawn. P(X=2)?

I do not know how to begin this, I'm really thrown off with "let X be value larger than 2 numbers drawn". Would that be like drawing a ball with 1 and a ball with 3 or a ball with 1 and a ball with a 2?

What steps would I need to do in order to obtain the solution of: P(x=2)=?

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    $\begingroup$ I suspect it means the larger of the values chosen. Thus if you choose any of $(1,2),(2,1),(2,2)$ the larger of the values chosen is $2$. $\endgroup$ – lulu Oct 16 '17 at 19:52
  • $\begingroup$ I agree that it is a weird formulation. If this is an exam question, you might want to suggest to be more clear on written exams. $\endgroup$ – DWe1 Oct 16 '17 at 19:54
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    $\begingroup$ An equivalent formulation is let $X,Y$ be i.i.d. random variables uniform on $\{1,2,3\}$. Define, $Z=\max\{X,Y\}$, what is $\mathbb{P}(Z=2)$ $\endgroup$ – clark Oct 16 '17 at 19:59
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$S={(1,1),(1,2), (1,3) (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}$

Therefore the sample space contains 9 items/events.

We assume that each combinations has an equal chance of being chosen, so each combination has a $1/9$ chance of being chosen.

$Pr(X=2)=Pr((1,2), (2,1) or (2,2))=1/9+1/9+1/9=3/9=1/3$

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  • $\begingroup$ Hi! Thank you for answering! I am wondering how you got 3/9 $\endgroup$ – Gill Dave Oct 16 '17 at 20:37
  • $\begingroup$ @GillDave Of the nine possible outcomes in the sample space, three have largest element $2$. They are $(1, 2)$, $(2, 1)$, and $(2, 2)$. $\endgroup$ – N. F. Taussig Oct 16 '17 at 20:40
  • $\begingroup$ @GillDave I updated the answer to show more steps $\endgroup$ – Amanda R. Oct 16 '17 at 20:46
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Oh, bother! Amanda R just pointed out the balls are numbered 1 to 3, not 0 to 2! The computation is basically the same.

The largest of the two numbers will be 2 for (1, 2), (2 ,1), or (2, 2). That is three out of the total 3*3= 16 possible results drawing two balls. P(2)= 3/9.

(Although you didn't ask, 1 is the "larger" only for (1,1). The probability is P(1)= 1/9. 1 is the larger only for (1, 1). The probability is P(1)= 1/9.

Three is larger for (1, 3), (2, 3), (3, 1), (3, 2) and (3, 3). The probability P(3)= 5/9.

Of course, 3/9+ 1/9+ 5/9= 9/9= 1.)

Thanks Amanda R!

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    $\begingroup$ OP stated that there are 3 balls numbered 1-3 $\endgroup$ – Amanda R. Oct 16 '17 at 20:45
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Let $X_1,X_2$ be your draws. Then $X_1,X_2$ are i.i.d. uniform on $\{1,2,3\}$. Now you want to compute $\mathbb{P}(Z=2)$ where $Z=\max \{X_1,X_2\}$.

Generally, you have $\mathbb{P}(Z\leq t) =\mathbb{P}(X_1\leq t,X_2\leq t)=\mathbb{P}(X_1\leq t)\mathbb{P}(X_2\leq t) $. Thereofore, \begin{align} \mathbb{P}(Z=2) &= \mathbb{P}(Z\leq 2)- \mathbb{P}(Z\leq 1)\\ &=\frac{2}{3} \cdot\frac{2}{3} - \frac{1}{3} \cdot \frac{1}{3} \\ &= 1/3 \end{align}

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