To solve the problem, let's first summarize the total number of balls in the box:
- Red balls: 14
- White balls: 4
- Blue balls: 10
Now, we can calculate the total number of balls:
\[ \text{Total number of balls} = 14 , (\text{red}) + 4 , (\text{white}) + 10 , (\text{blue}) = 28 \]
(a) Probability that the ball is either blue or red
To find the probability that a ball is either blue or red, we first need to find the total number of blue and red balls:
\[ \text{Number of blue or red balls} = \text{Number of red balls} + \text{Number of blue balls} = 14 + 10 = 24 \]
Now we can calculate the probability:
\[ P(\text{blue or red}) = \frac{\text{Number of blue or red balls}}{\text{Total number of balls}} = \frac{24}{28} \]
To simplify this fraction, we can divide the numerator and denominator by 4:
\[ P(\text{blue or red}) = \frac{24 \div 4}{28 \div 4} = \frac{6}{7} \]
(b) Probability that the ball is neither red nor white
To find the probability that a ball is neither red nor white, we only consider the blue balls. The number of blue balls is 10, as there are 0 other colors that fit this description.
\[ P(\text{neither red nor white}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{10}{28} \]
Again, we can simplify this fraction by dividing the numerator and denominator by 2:
\[ P(\text{neither red nor white}) = \frac{10 \div 2}{28 \div 2} = \frac{5}{14} \]
Final Answers
(a) The probability that the ball is either blue or red is \(\frac{6}{7}\).
(b) The probability that the ball is neither red nor white is \(\frac{5}{14}\).