Question

One ball is drawn from a bag in which there is one ball with number 3 on it, six balls with numbers 7, and three balls with numbers 11 . Let x be a random variable that assumes the number that is on the selected ball. Fill in the missing part of the probability

X 3 7
P(X)
distribution table. Please use the decimals in your answer.

Answer 1

To find the probabilities, we first need to determine the total number of balls in the bag.

- There is **1 ball** with the number **3**.
- There are **6 balls** with the number **7**.
- There are **3 balls** with the number **11**.

The total number of balls is:

\[
1 + 6 + 3 = 10
\]

Now, we can find the probability of drawing each type of ball by dividing the number of balls of each type by the total number of balls.

1. **Probability of drawing the ball with number 3**:
\[
P(X = 3) = \frac{\text{Number of balls with 3}}{\text{Total number of balls}} = \frac{1}{10} = 0.1
\]

2. **Probability of drawing a ball with number 7**:
\[
P(X = 7) = \frac{\text{Number of balls with 7}}{\text{Total number of balls}} = \frac{6}{10} = 0.6
\]

3. **Probability of drawing a ball with number 11**:
\[
P(X = 11) = \frac{\text{Number of balls with 11}}{\text{Total number of balls}} = \frac{3}{10} = 0.3
\]

Now, we can summarize the probability distribution table:

\[
\begin{array}{|c|c|c|c|}
\hline
X & 3 & 7 & 11 \\
\hline
P(X) & 0.1 & 0.6 & 0.3 \\
\hline
\end{array}
\]

Thus, the complete probability distribution table is:

\[
\begin{array}{|c|c|c|c|}
\hline
X & 3 & 7 & 11 \\
\hline
P(X) & 0.1 & 0.6 & 0.3 \\
\hline
\end{array}
\]