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: Determine total permutations

The total number of ways to arrange 7 street names is \( 7! \) (since it's a permutation of 7 distinct items). But for the first three positions, we can think step - by - step. The probability that the first street is Birch: there is 1 favorable outcome (Birch) out of 7 possible tree types. So the probability for the first street is \( \frac{1}{7} \).

Step2: Probability for the second street

After choosing Birch for the first street, there are 6 remaining tree types. The probability that the second street is Willow is \( \frac{1}{6} \) (since we want Willow out of the remaining 6).

Step3: Probability for the third street

After choosing Birch and Willow for the first two streets, there are 5 remaining tree types. The probability that the third street is Cedar is \( \frac{1}{5} \).

Step4: Calculate \( P(A) \)

Since these are independent events (in the sense of sequential selection), we multiply the probabilities: \( P(A)=\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5}=\frac{1}{210} \)

For Event B:

Step1: Total number of ways to choose first 3 streets

The total number of ways to choose and arrange 3 streets out of 7 is given by the permutation formula \( P(n,r)=\frac{n!}{(n - r)!} \), where \( n = 7 \) and \( r=3 \). So \( P(7,3)=\frac{7!}{(7 - 3)!}=\frac{7!}{4!}=7\times6\times5 = 210 \).

Step2: Number of favorable arrangements for Event B

The number of ways to arrange Cedar, Maple, and Oak without regard to order (i.e., the number of permutations of 3 items) is \( 3! = 6 \) (since for the first three positions, if we consider the set {Cedar, Maple, Oak}, the number of ways to arrange them is \( 3! \)).

Step3: Calculate \( P(B) \)

The probability \( P(B)=\frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements of first 3 streets}}=\frac{3!}{P(7,3)}=\frac{6}{210}=\frac{1}{35} \) (we can also think of it as: the probability that the first street is one of the three (Cedar, Maple, Oak) is \( \frac{3}{7} \), then the second street is one of the remaining two is \( \frac{2}{6} \), and the third street is the last one is \( \frac{1}{5} \). So \( P(B)=\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5}=\frac{6}{210}=\frac{1}{35} \))

Answer:

\( P(A)=\frac{1}{210} \)
\( P(B)=\frac{1}{35} \)