# Binomial Distribution

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I have a question and im stuck have been asked to find the probability of losing at least once and exactly once I'll post my question with my answers down below

Appreciate any help

Appreciate any help

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#3

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6iii)

**A0W0N**)6iii)

Exactly once means .

So for part (iii) you seek ... which can either be done by computing , or taking the shortcut by considering the fact that all .

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(Original post by

At least once means .

Exactly once means .

So for part (iii) you seek ... which can either be done by computing , or taking the shortcut by considering the fact that all .

**RDKGames**)At least once means .

Exactly once means .

So for part (iii) you seek ... which can either be done by computing , or taking the shortcut by considering the fact that all .

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#5

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Can you explain the short method i dont understand how to work out x from that

**A0W0N**)Can you explain the short method i dont understand how to work out x from that

You can easily work out what each of are and just add them up for .

The reason I say there is a shortcut is because all probabilities add up to 1:

and this is equivalent to

So just rearrange to get

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(Original post by

You're not supposed to 'work out X' it's a random variable which takes on values 0,1,...,6 and since it's binomially distributed you are able to compute the probabilities of it taking on any of these values.

You can easily work out what each of are and just add them up for .

The reason I say there is a shortcut is because all probabilities add up to 1:

and this is equivalent to

So just rearrange to get

**RDKGames**)You're not supposed to 'work out X' it's a random variable which takes on values 0,1,...,6 and since it's binomially distributed you are able to compute the probabilities of it taking on any of these values.

You can easily work out what each of are and just add them up for .

The reason I say there is a shortcut is because all probabilities add up to 1:

and this is equivalent to

So just rearrange to get

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I have another question this is to do woth significance level, i just dont know which values to assign n and p to ill attach my working and the question

2ii)

im not sure if ive done the the X~B part right so i stopped

2ii)

im not sure if ive done the the X~B part right so i stopped

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#8

(Original post by

I have another question this is to do woth significance level, i just dont know which values to assign n and p to ill attach my working and the question

2ii)

im not sure if ive done the the X~B part right so i stopped

**A0W0N**)I have another question this is to do woth significance level, i just dont know which values to assign n and p to ill attach my working and the question

2ii)

im not sure if ive done the the X~B part right so i stopped

Under we have that the number of times he rolls a 1 can be modelled by a random variable .

The p-value is just going to represent the probability of rolling a ONE exactly 5 times out of 12. I.e. P(X=5)

Then for the last part you need to compare this to the significance level in order to say what the conclusion should be.

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Im sorry to keep asking but i dont know what i should set H(null) and H(one) to for this question and even if i did which year i would choose

6i)

There is my working although i think it is completely wrong

6i)

There is my working although i think it is completely wrong

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#10

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Im sorry to keep asking but i dont know what i should set H(null) and H(one) to for this question and even if i did which year i would choose

6i)

There is my working although i think it is completely wrong

**A0W0N**)Im sorry to keep asking but i dont know what i should set H(null) and H(one) to for this question and even if i did which year i would choose

6i)

There is my working although i think it is completely wrong

The usual evidence over the history of the school shows that 2 out of 3 pupils achieve grades A-C. Since this is the accepted claim, it constitutes the null hypothesis.

Then a new tutor team steps in before a new academic year. The head of 6th form claims that this new tutor team will have an impact on the results. So we could get more pupils getting A-C grades, or less pupils. This constitutes a two-tailed alternate hypothesis.

After the year, it turns out that 16 out of 20 pupils got A-C grades. This is your sample and test statistic.

The head of maths says that this is an improvement due to a new teacher. An improvement claim means greater proportion of pupils getting A-C grades. Hence this implies an upper-tail test.

In your working out you should clearly define to be the proportion of students achieving A-C grades, and define X to be your random variable distributed binomially which represents the number of students that obtain A-C grade.

Last edited by RDKGames; 1 year ago

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#12

(Original post by

it gives me an answer above one

**A0W0N**)it gives me an answer above one

Last edited by RDKGames; 1 year ago

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(Original post by

Check what you wrote down... 2/3 appears TWICE ??

**RDKGames**)Check what you wrote down... 2/3 appears TWICE ??

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