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Binomial probability (1 Viewer)

deswa1

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Hey guys,

Can one of you help me with this binomial probability question (from Cambridge). I don't think you need to do the whole thing, I should be able to apply the method from i) to the other two parts. Thanks again :). The first part of the question was just to expand the identity I showed so I just did it to make it easier.

<a href="http://www.codecogs.com/eqnedit.php?latex=1)(a@plus;b@plus;c)^3=a^3@plus;b^3@plus;c^3@plus;3a^2b@plus;3a^2c@plus;3ab^2@plus;3ac^2@plus;3b^2c@plus;3bc^2@plus;6abc\\\\ \textup{Q: In a survey of 100 football supporters, 65 supported Hawthorn, 24 supported }\\ \textup{Collingwood and 11 followed Sydney (generalise this as a percentage- couldn't}\\ \textup{find the percentage button on this lol). Use the above expansion to find}\\ \textup{the probability that if three people are randomly selected:}\\ \textup{i) One supports Hawthorn, one supports Collingwood and one Supports Sydney}\\ \textup{ii) exactly two of them support Collingwood}\\ \textup{iii) at least two of them support the same team}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1)(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3ac^2+3b^2c+3bc^2+6abc\\\\ \textup{Q: In a survey of 100 football supporters, 65 supported Hawthorn, 24 supported }\\ \textup{Collingwood and 11 followed Sydney (generalise this as a percentage- couldn't}\\ \textup{find the percentage button on this lol). Use the above expansion to find}\\ \textup{the probability that if three people are randomly selected:}\\ \textup{i) One supports Hawthorn, one supports Collingwood and one Supports Sydney}\\ \textup{ii) exactly two of them support Collingwood}\\ \textup{iii) at least two of them support the same team}" title="1)(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3ac^2+3b^2c+3bc^2+6abc\\\\ \textup{Q: In a survey of 100 football supporters, 65 supported Hawthorn, 24 supported }\\ \textup{Collingwood and 11 followed Sydney (generalise this as a percentage- couldn't}\\ \textup{find the percentage button on this lol). Use the above expansion to find}\\ \textup{the probability that if three people are randomly selected:}\\ \textup{i) One supports Hawthorn, one supports Collingwood and one Supports Sydney}\\ \textup{ii) exactly two of them support Collingwood}\\ \textup{iii) at least two of them support the same team}" /></a>
 

barbernator

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wouldn't i) just be 6 x 0.65 x 0.24 x 0.11 from taking the term 6abc
 

deswa1

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wouldn't i) just be 6 x 0.65 x 0.24 x 0.11 from taking the term 6abc
Yep haha. Oh I get how to do these now. Thanks heaps Barb.

I love it how I find these so hard and then you do it easily and I realise I'm a retard...
 

barbernator

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Yep haha. Oh I get how to do these now. Thanks heaps Barb.

I love it how I find these so hard and then you do it easily and I realise I'm a retard...
haha everyone has their strong and weak points lol. If only there was such a skill to have a strong point as question 8's in their entirety
 

Sanjeet

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lol wow, could we be asked something like this in the HSC?
 

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