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Science & Mathematics Probability

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Bugz

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Another maths question :(

If the probability of me kicking a goal is 0.12 and I have 8 shots, what is the probability I will kick one?

And what is probability I will kick exactly two?
 
Somewhere between not likely and likely with 50/50 in the middle. :p
 
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
 

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its 50/50

you either kick it or you dont

simple really
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!
 
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%
 
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 ;)
 
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 :D:thumbsu::)
 

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