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What is the probability distribution for X?

A box has five tickets numbered 1, 2, 3, 4, and 5. Two are to be randomly selected without replacement. Let X be the number of occurrences of either 4 or 5. For example, if 2 and 5 are selected, then X=1. If both 4 and 5 are selected, then X=2.

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Explanation

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Explanation:

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Mar 20, 2018

$P \left(x\right) = \left\{\begin{matrix}\frac{1}{10} & x = 2 \\ \frac{6}{10} & x = 1 \\ \frac{3}{10} & x = 0\end{matrix}\right.$

Explanation:

We are choosing 2 tickets from 5 tickets, therefore, the number of combinations are:

( (n),(k) ) = (n!)/(k!(n-k!)

where $n = 5$ and $k = 2$

( (5),(2) ) = (5!)/(2!(3!)) = 10

Counting all of the possible outcomes:

$P \left(2\right) = \left(4 , 5\right) = \frac{1}{10}$
$P \left(1\right) = \left(1 , 4\right) , \left(2 , 4\right) , \left(3 , 4\right) , \left(1 , 5\right) , \left(2 , 5\right) , \left(3 , 5\right) = \frac{6}{10}$
$P \left(0\right) = \left(1 , 2\right) , \left(1 , 3\right) , \left(2 , 3\right) = \frac{3}{10}$

Please observe that $P \left(2\right) + P \left(1\right) + P \left(0\right) = 1$; this is a good indicator that the distribution is valid:

$P \left(x\right) = \left\{\begin{matrix}\frac{1}{10} & x = 2 \\ \frac{6}{10} & x = 1 \\ \frac{3}{10} & x = 0\end{matrix}\right.$

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