Original post by 34908seikjFor A, you multiply not ADD because the probability of winning or not winning the second game depends on the outcome of the first game.
Instead of 0.4+0.4, it would be 0.4*0.4 = 0.16
Which you probably just wrote addition by mistake, given that you got the right answer!
Part B, You need, the probability he will win in game one * by the probability that he does not win game two.
Then you also need to probability he does not win a game * the probability that he wins a game. Then just add the two together to get your answer
P(W,NW) - Probability of winning then not winning. = 0.4*0.6 = 0.24
P(NW,W) - Probability of not winning then winning. 0.6*0.4=0.24
Then we add the two probability together since we want the probability he wins only one game, and there are only two paths, if you like (outcomes) that can lead to Neil only winning one game.
So, it's 0.24+0.24 = 0.48
2A) So it wants the probability that neither at school, the probability that Anna isn't at school is 0.4, and the probability that Bill isn't at school is 0.3
It's conditional probability, since the probability of Bill going to school (or not going to school) depends on what Anna does, therefore we multiply.
0.4*0.3 = 0.12 - Which is the final answer since that is the only path (outcome) in which both Anna and bill DON'T go to school.
2B) At least one of them goes to school.
So it wants to know the probability that what the probability of Anna going to school when Bill bill does not, and when Anna doesn't go to school, but Bill does.
(2 possible outcomes [paths])
P(A,DA) - Probability that Anna attends but Bill does not attend. = 0.6*0.3 = 0.18
P(DA,A) - Probability that Anna does not attend but Bill does attend = 0.4*0.7=0.28
Now we add the two possible outcomes together to get 0.18+0.28 = 0.46
(Which you got correct!)