Question

a landscaper is selecting two trees to plant. he has five to choose from. three of the five are deciduous and two are evergreen. what is the probability that he chooses trees of two different types? express your answer as a percent. \bigcirc 30% \bigcirc 40% \bigcirc 50% \bigcirc 60%

Explanation:

Step1: Calculate total number of ways to choose 2 trees from 5.

The formula for combinations is $C(n, k)=\frac{n!}{k!(n - k)!}$. Here, $n = 5$ and $k=2$. So $C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4\times3!}{2\times1\times3!}=10$.

Step2: Calculate number of ways to choose 1 deciduous and 1 evergreen tree.

Number of ways to choose 1 deciduous from 3 is $C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3\times2!}{1\times2!}=3$. Number of ways to choose 1 evergreen from 2 is $C(2,1)=\frac{2!}{1!(2 - 1)!}=\frac{2\times1!}{1\times1!}=2$. So total ways for different types is $3\times2 = 6$.

Step3: Calculate probability and convert to percent.

Probability $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{6}{10}=0.6$. Convert to percent: $0.6\times100 = 60\%$.

Answer:

60%