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a store has 20 brown, 15 gray, 15 blue, and 10 white shirts. 3 shirts a…

Question

a store has 20 brown, 15 gray, 15 blue, and 10 white shirts. 3 shirts are selected at random. what set of fractions represents the probability that each shirt that is selected is brown, without replacement? a. $\frac{20}{60}cdot\frac{19}{60}cdot\frac{18}{60}$ b. $\frac{20}{60}cdot\frac{19}{59}cdot\frac{18}{58}$ c. $\frac{20}{60}cdot\frac{20}{60}cdot\frac{20}{60}$ d. $\frac{20}{60}cdot\frac{20}{59}cdot\frac{20}{58}$

Explanation:

Step1: Calculate total shirts

The total number of shirts is $20 + 15+15 + 10=60$.

Step2: Probability of first - brown shirt

The probability of selecting a brown shirt first is $\frac{20}{60}$ since there are 20 brown shirts out of 60 total shirts.

Step3: Probability of second - brown shirt

After one brown shirt is selected (without replacement), there are 19 brown shirts left and 59 total shirts left. So the probability of selecting a brown shirt second is $\frac{19}{59}$.

Step4: Probability of third - brown shirt

After two brown shirts are selected (without replacement), there are 18 brown shirts left and 58 total shirts left. So the probability of selecting a brown shirt third is $\frac{18}{58}$.

Answer:

B. $\frac{20}{60}\cdot\frac{19}{59}\cdot\frac{18}{58}$