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Twenty-one telephones have just been received at an authorized service center. Seven of these telephones are cellular, seven are cordless, and the other seven are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 21 to establish the order in which they will be serviced.a. What is the probability that all the cordless phones are among the first fourteen to be serviced?b. What is the probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced?c. What is the probability that two phones of each type are among the first six serviced?

Respuesta :

vlatkostojanovic

Answer:

a) P=1/116280

b) P=143/38760

c) P=441/2584

Step-by-step explanation:

We have seven of these telephones are cellular, seven are cordless, and the other seven are corded phones.

a)  We calculate the number of possible combinations

{21}_C_{14}=\frac{21!}{14! · (21-14)!}=116280

The number of favorable combinations is 1.

Therefore, the probability is

P=1/116280

b) We calculate the number of possible combinations

{21}_C_{14}=\frac{21!}{14! · (21-14)!}=116280

We calculate  the number of favorable combinations  

{14}_C_{7}=\frac{14!}{7! · (14-7)!}=429

Therefore, the probability is

P=429/116280

P=143/38760

c) We calculate the number of possible combinations

{21}_C_{6}=\frac{21!}{6! · (21-6)!}=54264

We calculate  the number of favorable combinations  

{7}_C_{2} · {7}_C_{2} · {7}_C_{2} =

=\frac{7!}{2!·(7-2)!} · \frac{7!}{2!·(7-2)!}  · \frac{7!}{2!·(7-2)!}

=21 · 21 · 21=9261

Therefore, the probability is

P=9261/54264

P=441/2584