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yuradex [85]
3 years ago
6

Why might Razis method not be a good way to estimate the probability? My team won 18 of their last 20 matches, so the probabilit

y that they will win their next match is 18/20 = 90%
Mathematics
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

It is a really idealistic way to estimate probability.

It usually works well for random events where there is no restriction like:

"There are N elements in a set, the probability of randomly selecting any element is the same (so there is no restriction) meaning that for every single element, the probability is 1/N.

Now suppose that there are K elements (K < N) with a given characteristic, then the probability of randomly selecting one of the K elements out of the sett with N elements is equal to K times the probability of each element, this is:

P = K*(1/N) = K/N"

Then if a dice has 6 possible outputs, and we want to find the probability of getting a 3 or a 5, we need to find the quotient between the outcomes where we get a 3 or a 5 (2) and the total number of outcomes (6)

P = 2/6

Now let's go to the case in the problem:

We know that out of the last 20 games, your team won 18.

Then yes, your team won (18/20)*100% = 90% of their played games.

Does this mean tath the probability of winning the next game is 90%?

Well no, because this is not a random selection, there are a lot of other things needed to analyze, as the win rate of the other team (suppose that the other team won 14 of their last 20 games, then their win rate is (16/20)*100% = 70%

Does this mean that the probability where our team wins is 90%, and the probability of the enemy team winning is 70%?

Well no, this makes no sense at all, and this is why this method is not a good way to estimate probability in complex cases like this one.

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7) PG &amp; E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many
BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>
5 0
2 years ago
WHOEVER ANSWERS THIS CORRECTLY IN 3 MINS GETS BRAINLIEST!!
tiny-mole [99]

Answer:

Step-by-step explanation:

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8 0
2 years ago
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saw5 [17]

Answer:

33

Step-by-step explanation:

So I just put them in order....18,19,23,25,28,30,31,33,33,35,36

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3 years ago
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Triss [41]

Answer:

the answer i got was 1159.1

Step-by-step explanation:

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is that close to any of your answers?

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3 years ago
I don’t understand this
IrinaVladis [17]
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