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

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umairfeb221999

Answer:

a) The probability of getting all (five) cordless phones among the first 10 is 0.0839.

b) The probability of having exactly two types in the last five is 0.24975.

c) To pick a sample of size 6 which contains two of each type of phone. The probability of selecting 2 phones from each type is 0.199800.

Step-by-step explanation:

a)  The first ten serviced (without regard to order within the first ten) constitute a random sample without replacement of size 10 from 15 phones.

So, such samples are:                  (Using Combination  "C")

                  (15C10) [tex]=3003\\[/tex]

We now regard the phones as of just two types: cordless or not-cordless. The not-cordless phones are the cellular and regular phones, and there are 10 such phones.

So, The number of ways to pick 5 cordless and 5 non-cordless is

                       (5C5) [tex]*\\[/tex] (10C5) [tex]=252[/tex]

Thus the probability of getting all (five) cordless phones among the first 10 is

                                               [tex]\frac{252}{3003} =0.0839[/tex]

b)  Regard the last 5 phones (without regard to order within the last 5) as a sample with replacement from the 15 phones.There are (15C5) such samples.

Thus

There are (3C2)= 3 ways to choose the two types. For each such choice there are (10C5)= 252 possible ways to get a sample of size 5 from only these two types of phones.

also as the 2 of these samples contain all phones of same type so,

                                                 252-2 = 250

250 ways are to pick from these two types, having both type restricted.

then the probability will be

                                               [tex]\frac{3*(250)}{15C5} =0.249756[/tex]

c)  We want to the find the probability that two phones of each type are among the first six serviced. The first six serviced can be regarded, when ignoring the order among the first 6, as an unordered sample of size 6 from 15 objects. so, there are (15C6) samples.

And there are [tex](15C2)^3[/tex] ways to pick from each 3 types of phones.

Thus the probability of selecting such a sample is

                                              [tex]\frac{(15C2)^2}{(15C6)}=0.199800[/tex]

azikennamdi

Using the Counting Techniques, we can determine that the solutions are as follows:

A) The probability that all the cordless phones are among the first ten to be serviced is 0.084.

B) the probability that after servicing ten of the phones, phones of only two of the three types remain to be serviced is 0.25

C) The probability that two phones of each type are among the first six serviced is 0.20.

Please see the attached for the full calculation.

What is the Counting Technique?

This is a fundamental technique in the probability that is used when the events being considered are above 2 in number.

See the link below for more about Counting Techniques:
https://brainly.com/question/25934918