ihatePE
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how do I go about doing question d please?

also just to make sure my answers for
a) 0.6 for Vice President 1 to 3
b) 0.064
c) 0.216


are those correct?
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MarkFromWales
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I'll just comment on part (a). You need to find the probabilities of 0 vice presidents attending, of 1 vice president attending, of two vice presidents attending and of 3 vice presidents attending. I suggest you model this with a binomial probability distribution (because there is a fixed number of trials, there are two possible outcomes, attendance and non-attendance, and the probability of attending is constant and the question says the outcomes are independent). Let X be the number of VPs who attend, the distribution of X is B(3, 0.6) and you will calculate P(X=0), P(X=1), P(X=2) and P(X=3). Or perhaps they just want you to state that the distribution is B(3, 0.6).
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RDKGames
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(Original post by ihatePE)
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how do I go about doing question d please?

also just to make sure my answers for
a) 0.6 for Vice President 1 to 3
b) 0.064
c) 0.216


are those correct?
a) asks for the type of distribution X follows. What famous one is applicable here?

b) is fine

c) is fine

d) is looking for E(X) so the expectation. Im sure you know the formula for this so use it.
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ihatePE
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(Original post by RDKGames)
a) asks for the type of distribution X follows. What famous one is applicable here?

b) is fine

c) is fine

d) is looking for E(X) so the expectation. Im sure you know the formula for this so use it.
I am aware of the E(X) formula, but to confirm, do I do 0.6(1)+0.6(2)+0.6(3)?
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MarkFromWales
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(Original post by ihatePE)
how do I go about doing question d please?

also just to make sure my answers for
a) 0.6 for Vice President 1 to 3
b) 0.064
c) 0.216


are those correct?
b and c are correct. For d, you have probably learned a formula for the expected value of a binomial distribution.
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RDKGames
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(Original post by MarkFromWales)
I'll just comment on part (a). You need to find the probabilities of 0 vice presidents attending, of 1 vice president attending, of two vice presidents attending and of 3 vice presidents attending.
I thought this at first too however upon rereading the Q my belief is they just want him to state B(3,0.6) because listing all probabilities would just make parts b and c kinda redundant.
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RDKGames
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(Original post by ihatePE)
I am aware of the E(X) formula, but to confirm, do I do 0.6(1)+0.6(2)+0.6(3)?
No.

P(X=1) is not 0.6... refer to the binomial distribution.
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MarkFromWales
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(Original post by RDKGames)
I thought this at first two however upon rereading the Q my belief is they just want him to state B(3,0.6) because listing all probabilities would just make parts b and c kinda redundant.
Yes, I agree.
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MarkFromWales
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(Original post by ihatePE)
I am aware of the E(X) formula, but to confirm, do I do 0.6(1)+0.6(2)+0.6(3)?
You should be worried that your method gives an answer greater than 3 !
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ihatePE
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(Original post by RDKGames)
No.

P(X=1) is not 0.6... refer to the binomial distribution.
I got 0.6 *3 = 1.8?


also how would you find p(just one attending)?
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RDKGames
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(Original post by ihatePE)
I got 0.6 *3 = 1.8?


also how would you find p(just one attending)?
Oh boy im not sure why youre doing 0.6*3...

Just one attending is 0.6*0.4*0.4 (ie one attends and the two others dont) but we can choose which one attends in (3 choose 1) ways hence the probability is

P(X=1) = (3 choose 1)*0.6*0.4^2
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ihatePE
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(Original post by MarkFromWales)
You should be worried that your method gives an answer greater than 3 !
(Original post by RDKGames)
No.

P(X=1) is not 0.6... refer to the binomial distribution.
sorry im asking dumb questions, it has been a long time since I've done maths and I just need quick answers to bring it all back, because this module im doing isn't interested in the methods, so im just needing this so I can interpret data
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ihatePE
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(Original post by RDKGames)
Oh boy im not sure why youre doing 0.6*3...

Just one attending is 0.6*0.4*0.4 (ie one attends and the two others dont) but we can choose which one attends in (3 choose 1) ways hence the probability is

P(X=1) = (3 choose 1)*0.6*0.4^2
thank you, it has been a rough couple of weeks and my brain has given up
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