Mmathematics Statistics

Hello,
Can someone please help me with this question, I am stuck at it. Thank you.
Tim has 3 possible tennis partners to team up with to play doubles: Greg, Andre and Pete. His results show that when he plays with Greg they win 40% of their matches, when he plays with Andre they win 70% of their matches and with Pete the figure is 80%. Tim plays a quarter of his games with Greg and three- fifths of his games with Andre.
a) Draw a tree diagram to represent this information.
b) Find the probability that Tim loses a match.
c) Find the probability that Tim wins a match, given that he has not been playing with Greg.
d) Given that Tim loses a match, find the probability that he has been playing with Pete.

Have you done any of it? Please post your working and clarify where you're stuck.

I have no clue for this question, I dont know how to start it, there is four people. I am stuck really.

First identify the events :

Event 1 : Choose a doubles partner (3 possible outcomes)
Event 2 : Win/lose the match (2 possible outcomes)

The two events will make up the two columns of your probability tree. So the first column will have three branches (Greg, Andre and Pete) and for each of those branches you'll have a win and a lose branch in the second column.

Does that help? Have a go at drawing the tree and please post it here if you get stuck or want us to check it.

Thank you, so with Greg will be 0.4 win and 0.6 Lose
with Andrew O.7 win 0.3 Lose
with Pete 0.8 win and 0.2 lose.
I dont understand what to do next which is 'Tim plays a quarter of his games with Greg and three-fiths of his games with Andre.

Thank you, so with Greg will be 0.4 win and 0.6 Lose
with Andrew O.7 win 0.3 Lose
with Pete 0.8 win and 0.2 lose.
I dont understand what to do next which is 'Tim plays a quarter of his games with Greg and three-fiths of his games with Andre.

Have you tried drawing a tree?

What's the probability that a random game is chosen and it is with Tim? It must be 1/4. So your tree will have three branches in the first column with probabilities 1/4 for reg, 3/5 for Andre and you can work out the probability for Pete.[/QUOT Am I right?

Did you get my image?

Did you get my image?

Your tree is fine and b) is correct but c) isn't.

Here's a useful formula for conditional ( "given that" ) probability:

P(A given B) = P(A and B) / P(B)

So

P(Wins given that he has not been playing with Greg) =

= P(Not been playing with Greg and wins) / P(Not playing with Greg)


Does that help?

Do I understand it?
Part C)
P=(21/50+3/25)/ 9/25=3/2?

Nearly but I'm not sure where you got 9/25 from. You're looking for P(Not playing with Greg) i.e. what is the probability that a random game is chosen and it is not with Greg?

You should be able to tell your answer was wrong because 3/2 is greater than 1.

ohh yeah, I think that I started to understand this. Should be P(Wins)=(21/50+3/25)/0.75=18/25?
3/4=21/50+9/50+3/25+3/100 (Without Greg)

18/25 is correct for c).

I'm not sure what your second line is?

Ok thank you.
Any clue for d) please. Given that Tim loses a match, find the probability that he has been playing with Pete.

P( playing with Pete given that he loses ) = P( playing with Pete and loses ) / P(loses)

My answer is 4/3 so I know that it is wrong.
=(3/25+3/100+9/50+3/20)/9/25?
It gives 4/3.

I'm not sure where you got those numbers from.

For P( playing with Pete and loses ), you just need to multiply along the bottom tree path so it's 3/20 x 0.2.

For P(loses) you need to add all of the paths that result in Tim losing.