Question

A box contains 4 blue, 2 green and 3 red balls and another box contains 4 red, 3 green, and 5 blue balls. A ball is picked up at random from one of the boxes. What is the probability that the ball is blue?

• A. \(\frac{2}{9} \)
• B. \(\frac{5}{24} \)
• C. \(\frac{19}{72} \)
• D. \(\frac{29}{72} \)
• E. \(\frac{31}{72} \)

Answer 1

Answer: (E)

Step 1: Understand the problem.
We are given two boxes, each containing a different set of colored balls:
- Box 1 contains 4 blue, 2 green, and 3 red balls.
- Box 2 contains 5 blue, 3 green, and 4 red balls.
A ball is picked randomly from one of the two boxes, and we need to find the probability that the ball is blue.

Step 2: Determine the total number of balls in each box.
- In Box 1, the total number of balls is: \[ 4 \, (\text{blue}) + 2 \, (\text{green}) + 3 \, (\text{red}) = 9 \, \text{balls} \] - In Box 2, the total number of balls is: \[ 5 \, (\text{blue}) + 3 \, (\text{green}) + 4 \, (\text{red}) = 12 \, \text{balls} \]

Step 3: Probability of picking a blue ball from each box.
- The probability of picking a blue ball from Box 1 is: \[ P(\text{blue from Box 1}) = \frac{4}{9} \] - The probability of picking a blue ball from Box 2 is: \[ P(\text{blue from Box 2}) = \frac{5}{12} \]

Step 4: Use the law of total probability.
Since a ball is picked randomly from one of the two boxes, the probability of picking a ball from either box is equal, i.e., each box has a probability of \( \frac{1}{2} \) of being chosen.
Thus, the total probability of picking a blue ball is: \[ P(\text{blue}) = \frac{1}{2} \times P(\text{blue from Box 1}) + \frac{1}{2} \times P(\text{blue from Box 2}) \] Substituting the values: \[ P(\text{blue}) = \frac{1}{2} \times \frac{4}{9} + \frac{1}{2} \times \frac{5}{12} \] Simplifying: \[ P(\text{blue}) = \frac{4}{18} + \frac{5}{24} \] To add these fractions, we find a common denominator. The least common denominator of 18 and 24 is 72. Thus: \[ P(\text{blue}) = \frac{16}{72} + \frac{15}{72} = \frac{31}{72} \]

Step 5: Conclusion.
The probability that the ball is blue is \( \frac{31}{72} \).

Final Answer:
The correct answer is (E): \( \frac{31}{72} \).