How to solve this There is this problem from past exam papers that I am trying to solve and I can't any ideas?
In a bet between two runners the winner will be the one who wins the other 3 times. If the terrain is wet, the 1st runner has a 60% chance of winning while if it is dry then the chance of winning is 50%. The 1st race is held on dry terrain, the 2nd and the 3rd on wet, the 4th if conducted, is held on dry and the 5th if needed on dry. The matches are independent. Find the probability of the 1st runner:
1)to win the bet
2)to win the bet at exactly the 4th race
3)to win the bet before the 4th race
4)to win the 1st race , to lose the 2nd and 3rd  and wind the 4th and 5th races
 A: I would calculate each probability of winning race 1 Dry, 2 Wet, 3 Wet, 4 Dry, 5Dry :
Winning probabilities for runner 1 at race x are noted 'P' and are:
P(xD) for Proba of winning a dry run, P(xW) for wet run:

*

*P(xD) = 0,50 is proba of a dry race win

*P(xW) = 0,60 is proba of a wet race win

*1-P(xD) = 0,50 is proba of a dry race loss

*1-P(xD) = 0,40 is proba of a wet race loss

Some calculs for runner 1 winning, so he wins at least 3 races :

*

*P(1D,2W,3W) = 0,5 * 0,60 * 0,60 = 0,18 (answer Question 3)

*P(1D,4D,5D) = 0,5 * (1-0,60) * (1-0,60) * 0,50 * 0,50 = 0,02  (answer Question 4)

*P(1D,3W,4D) = ... = 0,06

*P(1D,3W,5D) = 0,03

*P(2W,3W,4D) = 0,09

*P(2W,4D,5D) = 0,03

*P(2W,3W,5D) = 0,045

*P(3W,4D,5D) = 0,03

*P(1D,2W,4D) = 0,06

*P(1D,2W,5D) = 0,03

*P(best of 5) = sum of all previous P = 0,575 (answer Q1)

For Q2:
sum all the probabilities of winning 3 times in the first 4 races, the 4th beeing a win:

*

*P(1D,3W,4D) = 0,06

*P(2W,3W,4D) = 0,09

*P(1D,2W,4D) = 0,06

*sum equal = 0,21 (answer Q2)

