Question

there are 7 streets to be named after 7 tree types. ash, birch, cedar, fir, maple, oak, and willow. a city planner randomly selects the street names from the list of 7 tree types. compute the probability of each of the following events. event a: the first street is birch, followed by willow and then cedar. event b: the first three streets are cedar, maple, and oak, without regard to order. write your answers as fractions in simplest form. p(a) = p(b) =

Explanation:

For Event A:

Step1: Probability first is Birch

There are 7 tree types, so probability first street is Birch is $\frac{1}{7}$.

Step2: Probability second is Willow

After choosing Birch, 6 remain, so probability second is Willow is $\frac{1}{6}$.

Step3: Probability third is Cedar

After choosing Birch and Willow, 5 remain, so probability third is Cedar is $\frac{1}{5}$.

Step4: Multiply probabilities

To find the probability of all three events, multiply the probabilities: $\frac{1}{7} \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{210}$.

Step1: Total permutations of first 3

The number of ways to arrange 3 tree types out of 7 is $P(7,3)=\frac{7!}{(7 - 3)!}=7\times6\times5 = 210$.

Step2: Favorable permutations

The number of ways to arrange Cedar, Maple, and Oak (without regard to order in the sense that we just want these three in any order) is the number of permutations of 3 elements, which is $3! = 6$.

Step3: Calculate probability

Probability is favorable over total, so $P(B)=\frac{6}{210}=\frac{1}{35}$. (We can also think of it as: first, probability first is one of the three: $\frac{3}{7}$, then second is one of the remaining two: $\frac{2}{6}$, then third is the last one: $\frac{1}{5}$. Multiplying these: $\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5}=\frac{6}{210}=\frac{1}{35}$)

Answer:

$P(A) = \frac{1}{210}$

For Event B: