9. A pot has 6 shiny balls and 14 matte balls. If a ball is removed from the pot without replacement, what is the probability that the next ball taken will not be matte? $$ 30.00 \% $$ $$ 26.32 \% $$ $$ 31.58 \% $$

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UpStudy AI · Accepted Answer

Let $$ S $$ be the event that a ball is shiny (not matte). The pot initially contains 6 shiny and 14 matte balls, hence a total of 20 balls.

When one ball is removed without replacement, there are two cases:

1. **Case 1 (First ball is shiny):**  
   The probability that the first ball is shiny is  
   $$
   P(S_1) = \frac{6}{20} = \frac{3}{10}
   $$  
   After removing a shiny ball, the remaining balls are 5 shiny and 14 matte (total $$ 19 $$ balls). So the probability that the next ball is shiny is  
   $$
   P(S_2 \mid S_1) = \frac{5}{19}
   $$

2. **Case 2 (First ball is matte):**  
   The probability that the first ball is matte is  
   $$
   P(M_1) = \frac{14}{20} = \frac{7}{10}
   $$  
   After removing a matte ball, the remaining balls are 6 shiny and 13 matte. Thus, the probability that the next ball is shiny is  
   $$
   P(S_2 \mid M_1) = \frac{6}{19}
   $$

Using the law of total probability, the overall probability that the next ball will be shiny is  
$$
P(S_2) = P(S_1) \cdot P(S_2 \mid S_1) + P(M_1) \cdot P(S_2 \mid M_1)
$$
Substitute the computed probabilities:
$$
P(S_2) = \frac{3}{10} \cdot \frac{5}{19} + \frac{7}{10} \cdot \frac{6}{19} = \frac{15}{190} + \frac{42}{190} = \frac{57}{190}
$$

Simplify the fraction if possible:
$$
\frac{57}{190} \approx 0.30 = 30\%
$$

Thus, the probability that the next ball taken will not be matte is  
$$
30.00\%
$$
The probability that the next ball taken will not be matte is 30.00%.

9. A pot has 6 shiny balls and 14 matte balls. If a ball is removed from the pot without replacement, what is the probability that the next ball taken will not be matte? \( 30.00 \% \) \( 26.32 \% \) \( 31.58 \% \)

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