An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" $(h)$ and "tails" $(t)$ which we write $h t h$, $t t t$, etc. For each outcome, let $N$ be the random varlable counting the number of heads in each outcome. For example, if the outcome is $h t h$, then $N$ ( $h$ th) $=2$. Suppose that the random variable $X$ is defined in terms of $N$ as follows: $X=N-N^{2}-3$. The values of $X$ are given in the table below. \begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline Outcome & $h t h$ & $h h t$ & $t t t$ & $t h t$ & $h h h$ & $h t t$ & $t t h$ & $t h h$ \\ \hline Value of $X$ & -5 & -5 & -3 & -3 & -9 & -3 & -3 & -5 \\ \hline \end{tabular} Calculate the probabilities $P(X=x)$ of the probability distribution of $X$. First, fill in the first row with the values of $X$. Then fill in the approprlate probabilities in the second row.

Step 1 ：An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write h t h, t t t, etc.

Step 2 ：For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is h t h, then N ( h th) =2. Suppose that the random variable X is defined in terms of N as follows: X=N-N^{2}-3. The values of X are given in the table below.

Step 3 ：\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline Outcome & h t h & h h t & t t t & t h t & h h h & h t t & t t h & t h h \\ \hline Value of X & -5 & -5 & -3 & -3 & -9 & -3 & -3 & -5 \\ \hline \end{tabular}

Step 4 ：The first step is to calculate the total number of outcomes. Since we are tossing a coin 3 times, and each toss has 2 possible outcomes (heads or tails), the total number of outcomes is \(2^3 = 8\).

Step 5 ：Next, we need to count the number of outcomes for each value of X. From the table, we can see that X can take on the values -9, -5, and -3.

Step 6 ：The value of X is -9 for 1 outcome (h h h), -5 for 3 outcomes (h t h, h h t, t h h), and -3 for 4 outcomes (t t t, t h t, h t t, t t h).

Step 7 ：The probability P(X=x) is then the number of outcomes for which X=x divided by the total number of outcomes.

Step 8 ：\[P(X=-9) = \frac{1}{8} = \boxed{0.125}\]

Step 9 ：\[P(X=-5) = \frac{3}{8} = \boxed{0.375}\]

Step 10 ：\[P(X=-3) = \frac{4}{8} = \boxed{0.5}\]