Exercise in probability: prob 3 players going to shoot a goal are: 1/6,1/4,1/3.

EXERCISE 01:
The probability of three players going to shoot a goal are: 1/6,1/4,1/3..
a/_ give the probability that the three players have a score at the same time.
b/_ if they score at the same time,give the probability that each one score first.

Sidahmed Righi said:

EXERCISE 01:
The probability of three players going to shoot a goal are: 1/6,1/4,1/3..
b/_ if they score at the same time,give the probability that each one score first.

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Say that \(\displaystyle a,~b,~\&~c\) are the players.
Here is a list of the ways they can score: \(\displaystyle abc,~acb,~bac,~bca,~cab,~\&~cba\).
How many ways are there? In how of those is \(\displaystyle b\) first?

Your "score at the same time" is confusing- especially in "if they score at the same time, give the probability that each one score first"! My first reaction was "if they "score at the same time", the can't be one scoring first!"

I think what you intend is that probabilities that they score in a game (or period or halfare 1/6, 1/4, and 1/3 and then you are asking for the probability that they all score in the same game (or period or half ...). That probability is (1/6)(1/4)(1/3)= 1/72. As pka said, there are 3!= 6 different orders in which they could score and each person scores first in 1/3 of them. However, since a is more likely to score than b or c, we cannot assume that all 6 of those are equally likely.

well ..i'm sorry if you didn't understand the question but ,,what i mean is the three players are absolutely independent that mean each player have a max number of attempt and most likely the third player have the best chance to score with a 33% chance,however what i mean score at the same time is the three players are going to shoot the goal one after one and each one must score within the allowed attempt he have ,,i tried to solve it like this:

A is the first player.......the probability that he score is : 1/6=16%.......the probability that he won't score is :1-1/6=83%.
B is the second player.......the probability that he score is : 1/4=25%.......the probability that he won't score is :1-1/4=75%.
C is third player.......the probability that he score is : 1/3=33%.......the probability that he won't score is :1-1/3=66%.
&:

the probability that A score and the others don't: [ 1/6x(1-1/4)*(1-1/3)]= 6/72
the probability that B score and the others don't: [ 1/4x(1-1/6)*(1-1/3)]= 10/72
the probability that C score and the others don't: [ 1/3x(1-1/4)*(1-1/6)]= 15/72

so: the probability that they score at the same time=6/72+10/72+15/72=31/72..


2/_ the probability that each score first is:
A: P(A)xP=1/6x31/72=31/432=7%
B: P(B)xP=1/4x31/72=31/288=10%
C: P(C)xP=1/3x31/72=31/216= 14%


i'm really beginner in probability & statics ,,so i think this is really confusing!!!