A box contains a penny, two nickels, and a dime. If two coins are selected randomly from the box, without replacement, and if X is the sum... A box contains a penny (1¢), two nickels (5¢), and a dime (10¢). If two coins are selected randomly from the box, without replacement, and if $X$ is the sum of the values of the two coins,


*

*What is the probability distribution table of $X$?


$$\begin{array}{|c|c|c|c|c|}\hline X & 6¢ & 10¢ & 11¢ & 15¢ \\ \hline f(x) & 2/6 & 1/6 & 1/6 & 2/6\\\hline\end{array}$$  


*What is the cumulative distribution function $F(x)$ of $X$?

The cumulative distribution function, $F(x)$ of $X$ is defined as:   $F(x) = P(X ≤ x)$  

So would that mean I just write:
$P(X ≤ 6) = 2/6$
$P(X ≤ 10) = 1/6$
$P(X ≤ 11) = 1/6$
$P(X ≤ 15) = 2/6$  
\begin{align*}
P(X \leq 6) & = P(X = 6)=2/6\\
P(X \leq 10) & = P(X = 6) + P(X = 10)=2/6+1/6=1/2\\
P(X \leq 11) & = P(X = 6) + P(X = 10) + P(X = 11)=1/2+1/6=2/3\\
P(X \leq 15) & = P(X = 6) + P(X = 10) + P(X = 11) + P(X = 15)=2/3+2/6=1
\end{align*}
 A: A good notation for the $CDF$ would be
$$ F_{X}(x)=  
\begin{cases} 
1 & x \geq 15 \\
\frac{4}{6} & 11 \leq x \lt 15 \\
\frac{3}{6} & 10 \leq x \lt 11 \\
\frac{2}{6} & 6 \leq x \lt 10 \\
0 & x \lt 6
\end{cases} $$  
A: Some thoughts to get you going:


*

*You're missing the case where you select two nickels (10 cents)

*The probability of getting something less or equal than 11 must be at least the probability of getting something less or equal than 6. The cumulative distribution is the sum of probabilities up to that point.

*Your idea for calculating exactly $P(X=11)$ is correct, but your values are incorrect, that's why you get that weird negative answer.
A: Your answer to the first question is correct.
For the second question, the cumulative distribution function (CDF) of a random variable $X$ at $x$ is found by finding the probability that $X \leq x$. Since there are only four possible values for $X$, 
\begin{align*}
P(X \leq 6) & = P(X = 6)\\
P(X \leq 10) & = P(X = 6) + P(X = 10)\\
P(X \leq 11) & = P(X = 6) + P(X = 10) + P(X = 11)\\
P(X \leq 15) & = P(X = 6) + P(X = 10) + P(X = 11) + P(X = 15)
\end{align*}
As a sanity check, note that $P(X \leq 15) = 1$ since all the possible values for $X$ are less than or equal to $15$. 
