A0W0N
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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
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A0W0N
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6iii)
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RDKGames
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(Original post by A0W0N)
6iii)
At least once means X \geq 1.

Exactly once means X=1.

So for part (iii) you seek P(X \geq 1) ... which can either be done by computing P(X=1) + \ldots + P(X=6), or taking the shortcut by considering the fact that all P(0 \leq X \leq 6) = 1.
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A0W0N
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(Original post by RDKGames)
At least once means X \geq 1.

Exactly once means X=1.

So for part (iii) you seek P(X \geq 1) ... which can either be done by computing P(X=1) + \ldots + P(X=6), or taking the shortcut by considering the fact that all P(0 \leq X \leq 6) = 1.
Can you explain the short method i dont understand how to work out x from that
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RDKGames
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(Original post by A0W0N)
Can you explain the short method i dont understand how to work out x from that
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 P(X=1), \ldots, P(X=6) are and just add them up for P(X \geq 1).

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

P(X=0) + P(X=1) + \ldots + P(X=6) = 1

and this is equivalent to

P(X=0) + P(X \geq 1) = 1

So just rearrange to get

P(X \geq 1) = 1-P(X=0)
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A0W0N
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(Original post by 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 P(X=1), \ldots, P(X=6) are and just add them up for P(X \geq 1).

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

P(X=0) + P(X=1) + \ldots + P(X=6) = 1

and this is equivalent to

P(X=0) + P(X \geq 1) = 1

So just rearrange to get

P(X \geq 1) = 1-P(X=0)
thankyou sooo much it just clicked i understand it now
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A0W0N
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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
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2ii)
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im not sure if ive done the the X~B part right so i stopped
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RDKGames
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(Original post by 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
This is fine so far.

Under H_0 we have that the number of times he rolls a 1 can be modelled by a random variable X \sim B(12,\frac{1}{6}).

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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A0W0N
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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
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6i)
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There is my working although i think it is completely wrong
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RDKGames
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(Original post by 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
You misunderstand the context.


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 p 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.
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A0W0N
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it gives me an answer above one
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RDKGames
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(Original post by A0W0N)
it gives me an answer above one
Check what you wrote down... 2/3 appears TWICE ??
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A0W0N
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(Original post by RDKGames)
Check what you wrote down... 2/3 appears TWICE ??
yep, thanks😅
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