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Science & Mathematics Probability
- Thread starter Bugz
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- Tagged users None
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- Jul 25, 2010
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Somewhere between not likely and likely with 50/50 in the middle. 
SJ
Premium Platinum
p = 0.12, n = 8
X ~ B(n, p)
Pr(X = x) = nCx*p^x*(1 - p)^(n - x)
X ~ B(8, 0.12)
Pr(X = 1) = 8C1*0.12^1*(1 - 0.12)^(8 - 1) = 8*0.12*0.88^7 = 0.3923
Pr(X = 2) = 8C2*0.12^2*(1 - 0.12)^(8 - 2) = 28*0.12^2*0.88^6 = 0.1872
X ~ B(n, p)
Pr(X = x) = nCx*p^x*(1 - p)^(n - x)
X ~ B(8, 0.12)
Pr(X = 1) = 8C1*0.12^1*(1 - 0.12)^(8 - 1) = 8*0.12*0.88^7 = 0.3923
Pr(X = 2) = 8C2*0.12^2*(1 - 0.12)^(8 - 2) = 28*0.12^2*0.88^6 = 0.1872
Somewhere between not likely and likely with 50/50 in the middle.
Half and half
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JuddsABlue
Norm Smith Medallist
its 50/50
you either kick it or you dont
simple really
you either kick it or you dont
simple really
SJ
Premium Platinum
Ok so you'd give me 2/1 that I could kick a goal from one metre out?
Come on, it's 50-50 - either I kick it or I don't!
Less than Kent Kingsley, more than 2011 Cam Mooney.
Shell
Premium Platinum
Hahaha my response was going to be: "depends. ARE you Richo in this scenario?"
I worked it out like this.
You have 8 shots at goal and you have an 88% chance of missing each individual shot.
So, the probability of missing ALL 8 shots is:
0.88 x 0.88 x 0.88 x 0.88 x 0.88 x 0.88 x 0.88 x 0.88 = 35.963%
Therefore the chance of kicking at least one is 64.037%
You have 8 shots at goal and you have an 88% chance of missing each individual shot.
So, the probability of missing ALL 8 shots is:
0.88 x 0.88 x 0.88 x 0.88 x 0.88 x 0.88 x 0.88 x 0.88 = 35.963%
Therefore the chance of kicking at least one is 64.037%
Yeah there is an obvious difference between =1 and >=1.
When I read the question I too thought it probably related more to >=1 as the second part specifically mentioned exactly two whereas the first part didn't.
Bugz should clarify with his/her maths teacher before getting posters to do his/her homework
When I read the question I too thought it probably related more to >=1 as the second part specifically mentioned exactly two whereas the first part didn't.
Bugz should clarify with his/her maths teacher before getting posters to do his/her homework
- May 5, 2006
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If you follow SJ's workings for X=1,2...8 and sum the 8 results you'll end up with Dan26's answer anyway, so you if you calculate the probability of kicking exactly one, exactly two and at least one then you can't go wrong...
- Thread starter
- #12
Yeah there is an obvious difference between =1 and >=1.
When I read the question I too thought it probably related more to >=1 as the second part specifically mentioned exactly two whereas the first part didn't.
Bugz should clarify with his/her maths teacher before getting posters to do his/her homework
Haha, I do Intro to Quantitative methods by distance, can't be bothered ringing the 'tutor'- especially when geniuses like yourself are just a click away
