Question

There are 10 acts on a talent show. An acrobat comedian dancer guitarist juggler magician pianist singer violinist and a whistler. A talent show host randomly schedules the 10 acts Compute the probability of the events Event A First three acts dancer singer the guitarist in any order Event B the comedian first guitarist second and pianist third
P(A)=
P(B)=

Answer 1

Answer:

The value of P (A) is 0.00833.

The value of P (B) is 0.00139.

Step-by-step explanation:

It is provided that there are 10 acts on a talent show.

The two events are defined as follows:

A: First three acts dancer, singer and the guitarist in any order

B: The comedian first guitarist second and pianist third

(1)

Compute the number of ways to select 3 acts from the 10 as follows:

${10\choose 3}=\frac{10!}{3!\cdot (10-3)!}=\frac{10!}{3!\cdot 7!}=120$

There are 120 ways to select 3 acts from the 10 and only 1 way to select a dancer, singer and the guitarist in any order.

Compute the probability of selecting a dancer, singer and the guitarist in any order as follows:

$P(A)=\frac{1}{120}=0.00833$

Thus, the value of P (A) is 0.00833.

(2)

Compute the number of ways to select 3 acts from the 10 (without replacement) and with order as follows:

$^{10}P_{3}=\frac{10!}{ (10-3)!}=\frac{10!}{ 7!}=720$

There are 720 ways to select 3 acts from the 10 with order and only 1 way to select the comedian first guitarist second and pianist third.

Compute the probability of selecting the comedian first guitarist second and pianist third as follows:

$P(B)=\frac{1}{720}=0.00139$

Thus, the value of P (B) is 0.00139.