UCAT Medentry QR Question

For Jack to win the whole game, he has to win the next 2 rounds in a row because Josh is leading by 2-1. For each round the probability of winning is 0.5 so for Jack to win the next two rounds, the probability is 0.5 x 0.5 = 0.25. We know that one of them is definitely going to win the game becuase there are no draws so if the probability that Jack wins the game is 0.25 then the probability that Josh wins the game is 1 - 0.25 = 0.75

i get it that way but not when they did josh winning + jack winning x josh winning the next? why did they include jack's winning? but thank you! and sorry to be annoying how would you then go about working out this question with the same logic?

I calculated it as if Jack wins and then took away that probability from 1 to get the probability that Josh wins. Their explanation works it out the other way (which I guess does answer the question more directly but I think is the longer way).

For Josh to win overall he only needs to win one more round so, either he wins the next round (p = 0.5) or Jack wins the next round and Josh the following (p0.5 x p0.5 = 0.25) so the probability of Josh winning is the sum of these two options (0.5 + 0.25 = 0.75).

For this next question, you work it out the same way but you just have to bear in mind that the probabilities change depending on which counters they're using. Jim is leading 2-1 so for him to win overall he either wins the next game with the black counters which is p0.4 or he loses the next game with the black counters and then wins the following game with the white counters (p0.6 x p0.6 = 0.36) so the probability of Jim winning overall is the sum of these two options (0.4 + 0.36 = 0.76). He has a 76% chance of winning overall is which slightly higher than 3 in 4 (75%).