# 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.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
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:

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

• 13 minutes ago
• 14 minutes ago
• 14 minutes ago
• 17 minutes ago
• A minute ago
• 3 minutes ago
• 4 minutes ago
• 5 minutes ago
• 6 minutes ago
• 11 minutes ago
• 13 minutes ago
• 14 minutes ago
• 14 minutes ago
• 17 minutes ago