Qualification > Math

statistics

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SkyPilotage:

--- Quote from: AboD on May 03, 2011, 12:36:19 pm ---true. winning one game can be found by:
[probability of losing x probability of winning] + [probability of winning x probability of losing]
he can win the second or first

--- End quote ---
My point is the the question already gave you the probability of winning one game which includes both winning first or second...

NotAbod:
yes let me explain and answer this.
wat r all the chances in this game?
1-win both
2-lose both
3-win first, lose second
4-lose first, win second
probability of winning exactly one is given: 0.3, this include chance 3 and 4
probability of losing both is given: 0.28
sum of probabilities always add up to 1
let probability of winning both = y
0.28 + 0.3 + y = 1
y = 1- (0.28 + 0.3)
y = 0.42
probability of winning both is 0.42

SkyPilotage:

--- Quote from: AboD on May 03, 2011, 12:44:26 pm ---yes let me explain and answer this.
wat r all the chances in this game?
1-win both
2-lose both
3-win first, lose second
4-lose first, win second
probability of winning at least one is given: 0.3, this include chance 3 and 4
probability of losing both is given: 0.28
sum of probabilities always add up to 1
let probability of winning both = y
0.28 + 0.3 + y = 1
y = 1- (0.28 + 0.3)
y = 0.42
probability of winning both is 0.42

--- End quote ---
Ahaha dude just scroll up and look at my answer and yours  ;) I got 0.42 and u got 0.09
P.S: 0.3 is the probability of winning EXACTLY ONE not atleast one...
 

NotAbod:
i didnt read the question good, so i gave a rong answer
didnt notice that u answered lol
problem solved
Right exactly one not atleast one, will edit it

SkyPilotage:

--- Quote from: AboD on May 03, 2011, 12:51:31 pm ---i didnt read the question good, so i gave a rong answer
didnt notice that u answered lol
problem solved
Right exactly one not atleast one, will edit it

--- End quote ---
(sigh) Finally!!  ::)

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