Let's solve this problem step-by-step.
First, understand that when a coin is tossed two times, the possible outcomes are [tex]\( S = \{HH, HT, TH, TT\} \)[/tex], where:
- [tex]\( H \)[/tex] represents heads.
- [tex]\( T \)[/tex] represents tails.
Next, we want to find the probabilities for the number of times heads occurs ([tex]\( X \)[/tex]):
1. Number of Heads ([tex]\( X = 0 \)[/tex]):
- Only one outcome has zero heads: [tex]\( TT \)[/tex].
- Probability [tex]\( P_X(0) = a \)[/tex].
2. Number of Heads ([tex]\( X = 1 \)[/tex]):
- Two outcomes have one head: [tex]\( HT \)[/tex] and [tex]\( TH \)[/tex].
- Probability [tex]\( P_X(1) = b \)[/tex].
3. Number of Heads ([tex]\( X = 2 \)[/tex]):
- Only one outcome has two heads: [tex]\( HH \)[/tex].
- Probability [tex]\( P_X(2) = c \)[/tex].
Now let's fill out the probabilities:
1. Total Outcomes: There are 4 possible outcomes (HH, HT, TH, TT).
2. Probability Calculations:
- For [tex]\( X = 0 \)[/tex] (0 heads): There is 1 outcome (TT). The probability is [tex]\( \frac{1}{4} \)[/tex].
- For [tex]\( X = 1 \)[/tex] (1 head): There are 2 outcomes (HT, TH). The probability is [tex]\( \frac{2}{4} = \frac{1}{2} \)[/tex].
- For [tex]\( X = 2 \)[/tex] (2 heads): There is 1 outcome (HH). The probability is [tex]\( \frac{1}{4} \)[/tex].
So:
[tex]\[ a = 0.25 \][/tex]
[tex]\[ b = 0.5 \][/tex]
[tex]\[ c = 0.25 \][/tex]
Completed table:
[tex]\[
\begin{array}{|c|c|}
\hline
\multicolumn{2}{|c|}{\text{Coin Toss}} \\
\hline
\text{Heads: } X & \text{Probability: } P_X(x) \\
\hline
0 & 0.25 \\
\hline
1 & 0.5 \\
\hline
2 & 0.25 \\
\hline
\end{array}
\][/tex]
To summarize:
- [tex]\( a = 0.25 \)[/tex]
- [tex]\( b = 0.5 \)[/tex]
- [tex]\( c = 0.25 \)[/tex]
DONE
