Two chips drawn from urn, calculate $P(|\hat\theta - 3| > 1.0)$.

Question

Two chips are drawn without replacements from an urn containing 5 chips, numbered 1-5. The average of the two drawn is to be used as an estimator, $\hat\theta$, for the true average of all the chips ($\theta$). Calculate $P(|\hat\theta - 3| > 1.0)$.

solution: $P(|\hat\theta - 3| > 1.0) = P(\hat\theta < 2) + P(\hat\theta > 4) = P(\hat\theta = 1.5) + P(\hat\theta = 4.5) = P(1,2) + P(4.5) = \frac{2}{10}$.

I don't understand why the average is between $1.5$ and $4.5$

can someone please help me understand this problem? I understand until $P(|\hat\theta - 3| > 1.0) = P(\hat\theta < 2) + P(\hat\theta > 4) $ The answer is 2/10.

i would appreciate. Thanks

Answer 1

I don't understand why the average is between 1.5 and 4.5:

• The smallest two values that you can choose are $1$ and $2$, with an average of $1.5$.
• The largest two values that you can choose are $4$ and $5$, with an average of $4.5$.

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As to the question itself:

Let $X$ denote the average of the selected numbers

Then $P(|X-3|>1)=P([X<2]\vee[X>4])$

There are $\color{red}{10}$ ways to choose $2$ out of $5$ numbers:

• $1,2\implies{X=1.5<2}$
• $1,3\implies{X=2.0}$
• $1,4\implies{X=2.5}$
• $1,5\implies{X=3.0}$
• $2,3\implies{X=2.5}$
• $2,4\implies{X=3.0}$
• $2,5\implies{X=3.5}$
• $3,4\implies{X=3.5}$
• $3,5\implies{X=4.0}$
• $4,5\implies{X=4.5>4}$

As you can see, there are $\color{red}{2}$ cases where $[X<2]$ or $[X>4]$, hence the probability is $\color{red}{\dfrac{2}{10}}$.