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bixtya [17]
4 years ago
10

A group of 55 friends meets for lunch. They greet each other by exchanging fist bumps. How many fist bumps are exchanged if each

friend must bump with each of the 54 ​others? The total number of fist bumps exchanged is?
Mathematics
2 answers:
ANTONII [103]4 years ago
7 0
55 * 54 = 2,970, hope this helps.
Afina-wow [57]4 years ago
6 0
Well, there's a question I haven't really heard of before. Whatever the case, the total number of fist bumps exchanged is 2,970. (That's a lot of people. That's even more fist bumps. Jeesh.)

Here's how I got that answer : each person exchanges a fist bump with 54 others. So that means each person fist bumps a total of 54 times each. And there are 55 friends in that group, which would mean that you would have to multiply.

54 × 55 = 2,970
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Answer:

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Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

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a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990

c) The probability of more than one case is:

P(k>1)=1-P(k\leq 1)=1-0.990=0.010

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

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