Thread: Probability?

Ok, on "c) Probability that there won't be an away victory?" I figured :
P(H) = 0,45 + P(U) = 0,29 divided by 1.

Does that add up?

Also, I'm not sure how to approach these questions, could anyone give any kind of hints?

Or a good explained answer?

Originally Posted by Rawr

Right, so I've got an upcoming exam tomorrow, and I just found out I'm stuck on probabilities (Not sure if that is what it's called in English).

Anyways, the question goes like this :

Troms skiing facility got 6 menn for NM in skiing. 2 of them are under the age of 20. It's a 4x7,5km lap, and they are going to pick 4 "runners(/skiiers)".

a) What's the probablity for that NONE of the 2 under the age of 20 does this?
b) What's the probablity for that minimum ONE of the 2 under the age of 20 does this?
c) What's the probablity for that EXACTLY ONE of the 2 under the age of 20 does this?
d)What's the probability that the skiier hit at least 4 of 5 targets with 5 shots.? (Hitting 1 target = 90%)

If you could explain your answers, that would be great.
I will most likely understand it, if you just set it up correctly.

[snip]

a) .


b) .


c) .


d) X ~ Binomial(n = 5, p = 0.9). Calculate .

Originally Posted by Rawr

[snip]
Statistic over football matches in 2005 showed the probabilities for home victiory (H), tie (T), away victory (A).
P(H) = 0,45, P(U) = 0,29, P(B)= 0,26 Mr F says: What events do U and B represent?

a) What's the probability that ALL the twelve matches ends with home victory?
b) How big is the posibility that the 6 first ends with home victory, and the next tree tie, and the last tree with away victory?
c) Probability that there won't be an away victory?

a) (0,45)^12.

Thank you mr. fantastic, even though it was a day late.

I've already had my exam now .


I did however find out how to do it in time.
The formula I posted worked just fine, I think I accidentally wrote the wrong number or something when I tried it out.