Alright, let's tackle this question step-by-step.
1. Identifying the unique values of [tex]\( X \)[/tex]
From the data given:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline Outcome & t h h & h h t & t t h & h t t & h t h & t t t & h h h & t h t \\
\hline Value of $X$ & -4 & -4 & -3 & -3 & -4 & 0 & -3 & -3 \\
\hline
\end{tabular}
\][/tex]
The unique values of [tex]\( X \)[/tex] are [tex]\( -4 \)[/tex], [tex]\( -3 \)[/tex], and [tex]\( 0 \)[/tex].
2. Counting the occurrences for each value of [tex]\( X \)[/tex]
- For [tex]\( X = -4 \)[/tex]: The outcomes where [tex]\( X = -4 \)[/tex] are [tex]\( t h h \)[/tex], [tex]\( h h t \)[/tex], and [tex]\( h t h \)[/tex]. Hence, the count is [tex]\( 3 \)[/tex].
- For [tex]\( X = -3 \)[/tex]: The outcomes where [tex]\( X = -3 \)[/tex] are [tex]\( t t h \)[/tex], [tex]\( h t t \)[/tex], [tex]\( h h h \)[/tex], and [tex]\( t h t \)[/tex]. Hence, the count is [tex]\( 4 \)[/tex].
- For [tex]\( X = 0 \)[/tex]: The outcome where [tex]\( X = 0 \)[/tex] is [tex]\( t t t \)[/tex]. Hence, the count is [tex]\( 1 \)[/tex].
3. Calculating the probabilities
We have a total of [tex]\( 8 \)[/tex] possible outcomes (since a fair coin is tossed [tex]\( 3 \)[/tex] times, [tex]\( 2^3 = 8 \)[/tex]).
Now, we calculate the probabilities [tex]\( P(X=x) \)[/tex] as follows:
- [tex]\( P(X = -4) = \frac{3}{8} = 0.375 \)[/tex]
- [tex]\( P(X = -3) = \frac{4}{8} = 0.5 \)[/tex]
- [tex]\( P(X = 0) = \frac{1}{8} = 0.125 \)[/tex]
4. Filling in the table
Now, we can fill in these values into the table:
[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline Value $x$ of $X$ & -4 & -3 & 0 \\
\hline $P ( X = x )$ & 0.375 & 0.5 & 0.125 \\
\hline
\end{tabular}
\][/tex]
This completes our detailed solution.
