Question

last summer chloe caught 32 trout, 20 catfish, 11 salmon, and 9 small-mouth bass.

if chloe goes fishing again next summer, and the fish populations have not changed, what is the probability that the first fish she catches will be a small-mouth bass?

a.) \\(\frac{1}{9}\\)

b.) \\(\frac{1}{4}\\)

c.) \\(\frac{1}{8}\\)

d.) \\(\frac{1}{12}\\)

Explanation:

Calculate the total number of fish caught

To find the total number of fish in the sample, we sum the quantities of each type of fish caught by Chloe:

$$ \text{Total fish} = 32 \text{ (trout)} + 20 \text{ (catfish)} + 11 \text{ (salmon)} + 9 \text{ (small-mouth bass)} $$

$$ \text{Total fish} = 32 + 20 + 11 + 9 = 72 $$

Identify the number of favorable outcomes

The event of interest is catching a small-mouth bass. The number of small-mouth bass caught is:

$$ \text{Favorable outcomes} = 9 $$

Calculate the probability

The probability \(P\) of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes:

$$ P(\text{small-mouth bass}) = \frac{\text{Favorable outcomes}}{\text{Total fish}} = \frac{9}{72} $$

Simplify the fraction

We simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 9:

$$ P(\text{small-mouth bass}) = \frac{9 \div 9}{72 \div 9} = \frac{1}{8} $$

Answer:

• a.) \(\frac{1}{9}\)
• b.) \(\frac{1}{4}\)
• c.) \(\frac{1}{8}\) (Correct answer)
• d.) \(\frac{1}{12}\)