IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: sridhar on May 02, 2011, 07:52:04 pm
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Hi any body help with this probability quiz plz!
Paul plays 2 games. the probability he wins exactly one game is 0.3 and the probability he looses both games is 0.28.
what is the probability he wins both games?
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I think its either 0.3x0.3=0.09 or 1-0.28=0.72
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Thanks, but I am not sure but how about drawing a tree diagram
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2 things
1st-pro is right its 0.3x0.3
2nd- wrong section
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Is the answer (9/112) ?
By the way, moving this to AS/A2 section of math.
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Welcome back fidato :D
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Hmm i think its 1-0.28-0.3!!= 0.42
Because if he doesnt win both games he can either lose both or win only one ....
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he can:
win both
lose both
win first, lose second
lose first, win second
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he can:
win both
lose both
win first, lose second
lose first, win second
You are perfectly right, however the probability 0.3 is the probablity he wins EXACTLY one game and not the probablity to win any game.
Did you get what I mean?
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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
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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
My point is the the question already gave you the probability of winning one game which includes both winning first or second...
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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
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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
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...
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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
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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
(sigh) Finally!! ::)
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Welcome back fidato :D
Thanks.
You both - substitute the values and check, (i.e. sq.rt. of 0.42 and subtract the answer from 1 and hence square, this will give you the probability of loosing two in a row).
Get my point ?
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nope, didnt get it
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explain this point please
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I think question has faulty values.
Okay, study this :
If you don't understand, then tell me.
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if prob. of winning is 1-sq0.28
then winning both is [1-sq0.28] * [1-sq0.28] = 0.22 ?
not sure from this
chatroom?
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I think question has faulty values.
Okay, study this :
If you don't understand, then tell me.
Ahaha Fidato, Very good point :) but you would be correct if you are sure that tht probability is independant... but its not!!
This means that the prob of winning the first time is DIFFERENT than the probabiity of winning the second game!!!!
Got it? :) P.S: if you aren't convinced try finding the prob of winning exaclty one game it will come up as 0.5 not 0.3...
Life is simple, no need to go all through this ;)
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maybe that dude copied the wrong values
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maybe that dude copied the wrong values
Everything is possible but it wont matter ;D
Use the values you have...I am pretty sure the probabilities aren't the same in the first and second time :-\
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Ahaha Fidato, Very good point :) but you would be correct if you are sure that tht probability is independant... but its not!!
This means that the prob of winning the first time is DIFFERENT than the probabiity of winning the second game!!!!
Got it? :) P.S: if you aren't convinced try finding the prob of winning exaclty one game it will come up as 0.5 not 0.3...
Life is simple, no need to go all through this ;)
You came to my point. ;)
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Ahaha Fidato, Very good point :) but you would be correct if you are sure that tht probability is independant... but its not!!
This means that the prob of winning the first time is DIFFERENT than the probabiity of winning the second game!!!!Got it? :) P.S: if you aren't convinced try finding the prob of winning exaclty one game it will come up as 0.5 not 0.3...
Life is simple, no need to go all through this ;)
Think on this again. If probabilities of winning the first and second games would be different, then lossing both the games would not be (sq.rt. 0.28).
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Think on this again. If probabilities of winning the first and second games would be different, then lossing both the games would not be (sq.rt. 0.28).
You mean the probability of losing one game would not be sq.rt of 0.28? Exactly but why do we care ::)