5 women are in a race. What is the probability that Sally and Jenny both will not win the race? What is the probability that Sally, but not Jenny, will win the race?
probability Sally wins=0.2 probablility Jenny wins =0.2 probability neither wins 0.8*0.8=64% probabilty one wins=0.2 cause they exclude themselves ?
Is Sally faster than Jenny? 3/5 1/5 I suppose are the answers…
Assuming that there can be only one winner and that the chances of winning are equally distributed, here’s my reasoning … [1-P(Sally not winning)] * [1-P(Jenny not winning)] = 64%. //I think florinp is right Sally and not Jenny … we’re essentially removing Jenny from the list of possible winners. It’s as though she did not even enter the race. So it’s the chances of Sally winning in a group of four instead of five: 25%.
P(Sally not Jenny winning the race) = P(Sally winning the race) cause only one person can win the race.
zeroaffinity Joey is right they exclude themselves but not winning both of them isnot an mutually exclusive event
P(SL and JL)=P(SL)+P(JL)-P(SL or JL)=.8+.8-1=.6=3/5
Of course we have some fine Alice: “The race is over!” and they all crowded round it, panting, and asking, “But who has won?” This question the Dodo could not answer without a great deal of thought, and it sat for a long time with one finger pressed upon its forehead (the position in which you usually see Shakespeare, in the pictures of him), while the rest waited in silence. at last the Dodo said, “*Everybody* has won, and all must have prizes.”
I realised my mistake So if Sally loses p=0.8 there is 4 more contestants that can win probabilyty that Jenny loses=1-0.25=0.75 probability that both lose=0.8*0.75=0.6
>I realised my mistake >So if Sally loses p=0.8 >there is 4 more contestants that can win probabilyty that Jenny loses=1-0.25=0.75 >probability that both lose=0.8*0.75=0.6 D’oh. I need to bump my stats review up a couple of places on the things-to-not-screw-up list.
quants and probabilities are killing me… Joey is going to start laughing