Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cord

Question

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Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cord

less, and the other five are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, ……15 to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten 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?

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-02-28T20:38:06+03:00

Answer:

a) The likelihood of having all five cordless phones among the first 10 is 0.0839.

b) The chance of having exactly two types among the last five is 0.24975.

c) For a sample size of 6 that includes two phones of each type, the probability of selecting 2 of each kind is 0.199800.

Step-by-step explanation:

a) The first ten phones that are serviced, without considering their order, represent a random sample of size 10 from 15 total phones.

The potential samples can be represented as: (Utilizing combination "C")

((15C10) $=3003\\$ )

Considering the phones as belonging to two categories: cordless and non-cordless. There are 10 non-cordless phones, which are either cellular or corded.

The combinations available for choosing 5 cordless and 5 non-cordless phones are given by:

((5C5) $*\\$ (10C5) $=252$ )

Thus, the probability of all 5 cordless phones being in the first 10 is:

$\frac{252}{3003} =0.0839$

b) The last five phones can be viewed as a sample with replacement from the 15 phones. The number of such samples would be ((15C5)).

In this case, there are (3C2)=3 options for selecting two types. For each choice, there are (10C5)=252 ways to gather a sample of size 5 from just those two types of phones.

However, since 2 of these samples consist of all phones of a single type,

the total becomes 252-2 = 250

So, 250 ways exist to select from these two types while ensuring both types are present.

The probability becomes:

$\frac{3*(250)}{15C5} =0.249756$

c) We need to determine the probability that two phones from each type appear among the first six serviced. Ignoring order, these first six can be treated as an unordered sample of size 6 extracted from 15 items. This gives us (15C6) possible samples.

Furthermore, there are $(15C2)^3$ combinations available to choose from each of the three types of phones.