To determine the probability distribution for the number of heads in three tosses of a fair coin, let's break it down step by step.
1. Identify the possible outcomes: Each coin toss can result in either a head (H) or a tail (T). For three tosses, the number of possible outcomes is [tex]\(2^3 = 8\)[/tex]. The possible outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
2. Count the number of heads in each outcome:
- 0 heads: TTT (1 outcome)
- 1 head: HTT, THT, TTH (3 outcomes)
- 2 heads: HHT, HTH, THH (3 outcomes)
- 3 heads: HHH (1 outcome)
3. Calculate the probabilities:
- Probability of 0 heads ([tex]\(P(X=0)\)[/tex]):
[tex]\[
P(X=0) = \frac{\text{Number of outcomes with 0 heads}}{\text{Total number of outcomes}} = \frac{1}{8}
\][/tex]
- Probability of 1 head ([tex]\(P(X=1)\)[/tex]):
[tex]\[
P(X=1) = \frac{\text{Number of outcomes with 1 head}}{\text{Total number of outcomes}} = \frac{3}{8}
\][/tex]
- Probability of 2 heads ([tex]\(P(X=2)\)[/tex]):
[tex]\[
P(X=2) = \frac{\text{Number of outcomes with 2 heads}}{\text{Total number of outcomes}} = \frac{3}{8}
\][/tex]
- Probability of 3 heads ([tex]\(P(X=3)\)[/tex]):
[tex]\[
P(X=3) = \frac{\text{Number of outcomes with 3 heads}}{\text{Total number of outcomes}} = \frac{1}{8}
\][/tex]
4. Express the probabilities as a probability distribution table:
[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
x & 0 & 1 & 2 & 3 \\
\hline
P(X=x) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\
\hline
\end{array}
\][/tex]
So, the probability distribution for the number of heads in three tosses of a fair coin is:
[tex]\[
P(X=0) = \frac{1}{8}, \quad P(X=1) = \frac{3}{8}, \quad P(X=2) = \frac{3}{8}, \quad P(X=3) = \frac{1}{8}
\][/tex]
