1. ## Conditional probability

Rachel and annah play each other at badminton. each games results in either a win for Rachel or a win for Annah, the P of rach winning the first game is 0.6, If rach wins the first game the prob of her winning the next game is 0.7 but if she loses the P is 0.4

Find conditional prob that rach wins the first game given that she loses the second one....

so P (AUB) = wins 1st and loses second... = 0.6 x 0.3 = 0.18

And prob she wins the first game = 0.6

So the Conditional prob = 0.18/0.6 = 0.3 ?

Help ?

and second part, find P that rachel wins 2 games and loses 1 game out of the first free games she plays ?

the p of only 2 match is given ? :s all confused...

2. If I follow the question:
You are asked to find $\displaystyle P(R_1|R_2^{~c})$.

Now $\displaystyle P(R_2^{~c})=P(R_1R_2^{~c})+P(R_1^{~c}R_2^{~c})=(.6 )(.3)+(.4)(.4)$.

3. The first part the answer is 0.429, and the way i learnt conditional probability is, P(A U B)/A....

And for the second part, that leaves me some choices i think, like she wins first, then loses second and wins third,
and etc etc, so could be a combination ?

4. Originally Posted by ramzel
The first part the answer is 0.429, and the way i learnt conditional probability is, P(A U B)/A....
First of all, I disagree with that given answer.
The answer I get is $\displaystyle \color{blue}0.529$ so I think that there is a typo in your text: the 4 should be 5.

Next I do not think that you fully understand the notation of conditional probability.
It involves intersection not union: $\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$

5. I got it the answer i got is 0.429,

P she loses second game = 0.18 + 0.24

P she wins first and loses second = 0.18

0.18/0.42 = 0.42857

thank you Plato for your help,

And the second part ? how do i do it

6. ok