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

Question

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:

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