quick probability question

blacklionboy

go this question as a tutorial question.

At the start of the season a Premiership footballer has probability 0.9 of scoring a penalty.

"""""
(a) If his team is awarded 4 penalties over the course of a season, what is the probability that he scores all four of the penalties? What assumptions have you made?
(b) What is the probability that he scores 3 and misses 1 of the four penalties?
(c) How many penalties is the expected value of the number of penal- ties taken before he scores one?""""

just wanna check my answers with everyone cause ive no access to solutions.

a) (0.9)^4=0.6561
b) (0.9)^3 X (0.1) = 0.0729
c) this one was was wierd. I said P(misses 1st)=0.1, P(misses 2nd)=0.01, P(misses 3rd)=0.01 ect.
Then as E(x)= Sum(x p(x)) = 1(0.1)+2(0.01)+3(0.001)=0.12345

So he should take 0.12345 penos till scoring.



Mainly part C im wondering abour. any takers? thanks!

Michael Collins

a) This is correct - what is your main assumption here?
b) Almost correct - you have calculated that he scores the first three and misses the last one, are there other possibilites that you can think of?
c) Your answer here does not quite make sense, it suggests he scores the penalty, on average, before he even takes one! Have you learnt about the geometric distribution? The anwer is in here: http://en.wikipedia.org/wiki/Geometric_distribution.

Post back if ya get stuck or have answers you want checked.