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Ray Of Light [21]
3 years ago
10

A group of five people are all working on the same mathematics problem. On the night before it is due, they call each other to d

iscuss their work. Each person talks to all the other people at least once. What is the fewest number of telephone calls that could be made?
Mathematics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

Minimum number of calls = 10

Step-by-step explanation:

Lets name the five people as A,B,C,D and E.

<u>On the night before ,each person talks to every other person atleast once that means A would talk to B,C,D and E atleast once.</u>

Lets start with A. He would talk to other 4 people which means there would be 4 phone calls made.

Now lets take B. He can talk to A,C,D and E. But  A has already talked to C therefore to get minimum number of phone calls , B need not call A again. So he calls only C,D and E.

In case of C using similar logic he need to talk to only D and E.

For D , he talks to E alone.

E does not have to talk to anyone as he has already talked to everyone atleast once.

Total calls = 4 + 3 + 2 + 1

                    = 10

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Answer:

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(b) The probability that at least one of them will show up is 0.75.

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Step-by-step explanation:

Denote the events as follows:

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<em>M</em> =  Mike will show up.

Given:

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(a)

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(b)

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P (At least one of them will show up) = 1 - P (Neither will show up)

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(c)

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P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

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