Question

there are 7 acts in a talent show. an acrobat, a comedian, a dancer, a guitarist, a juggler, a singer, and a whistler. a talent show host randomly schedules the 7 acts. compute the probability of each of the following events. event a: the juggler is first, the whistler is second, and the singer is third. event b: the first three acts are the guitarist, the whistler, and the singer, in any order. write your answers as fractions in simplest form. p(a) = □ p(b) = □

Explanation:

Step1: Calculate \(P(A)\)

The total number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 7\), so the total number of ways to schedule the \(7\) acts is \(7! = 7\times6\times5\times4\times3\times2\times1=5040\).
For event \(A\): The juggler is first, the whistler is second, and the singer is third. The remaining \(7 - 3=4\) acts can be arranged in \(4!\) ways.

$$P(A)=\frac{4!}{7!}=\frac{4\times3\times2\times1}{7\times6\times5\times4\times3\times2\times1}=\frac{1}{7\times6\times5}=\frac{1}{210}$$

Step2: Calculate \(P(B)\)

The number of ways to choose \(3\) acts out of \(7\) for the first - three positions is given by the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 7\) and \(r = 3\), \(P(7,3)=\frac{7!}{(7 - 3)!}=\frac{7!}{4!}=7\times6\times5 = 210\).
The number of ways to arrange the remaining \(7-3 = 4\) acts is \(4!\). The number of favorable arrangements for event \(B\) (the first three acts are the guitarist, the whistler, and the singer in any order) is \(3!\times4!\) ( \(3!\) for arranging the three specific acts among the first three positions and \(4!\) for arranging the remaining four acts).

$$P(B)=\frac{3!\times4!}{7!}=\frac{3\times2\times1\times4!}{7\times6\times5\times4!}=\frac{6}{210}=\frac{1}{35}$$

Answer:

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