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25 Aug 2010, 19:22
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:25) correct
37%
(02:12)
wrong
based on 46
sessions
History
Date
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A bag contains 6 red marbles,9 blue marbles,and 5 green marbles.You withdraw one marble,replace it,and then withdraw another marble. What is the probability that you do not pick two green marbles?
A. 3/4
B. 1/2
C. 8/6
D. 5/12
E. 10/80
probability that you do not pick two green marbles
P(g and g) = 5/20 = 1/4
P(g and g) = 5/20 = 1/4
P(not two green) = 1-1/4 = 3/4
I know this is a silly doubt..please...can someone clarify it...thankyou in advance...
probability of not happening=1-p(happening)
P(not G)=1-P(G)
here,calculating P(G) for the first attempt=5/20
and because the marble is replaced and P(G)=5/20
then according to me,as there are two events, P(GandG)=2(5/20)=1/2
and then a/c to formula =1-1/2=1/2
but i know that it is not 1/2
can someone please give me an explanation.i know these are the basics of probability.. im really sorry to ask such a silly doubt
A. 3/4
B. 1/2
C. 8/6
D. 5/12
E. 10/80
probability that you do not pick two green marbles
P(g and g) = 5/20 = 1/4
P(g and g) = 5/20 = 1/4
P(not two green) = 1-1/4 = 3/4
I know this is a silly doubt..please...can someone clarify it...thankyou in advance...
probability of not happening=1-p(happening)
P(not G)=1-P(G)
here,calculating P(G) for the first attempt=5/20
and because the marble is replaced and P(G)=5/20
then according to me,as there are two events, P(GandG)=2(5/20)=1/2
and then a/c to formula =1-1/2=1/2
but i know that it is not 1/2
can someone please give me an explanation.i know these are the basics of probability.. im really sorry to ask such a silly doubt
This Question is Locked Due to Poor Quality
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25 Aug 2010, 20:13
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harithakishore
I don' think that there is a correct answer among answer choices. Can you please post the source of this question and OE.
We are asked to find the probability that we won't have scenario GG.
The probability of this event is \(P(GG)=\frac{5}{20}*\frac{5}{20}=\frac{1}{16}\) and the probability of this event NOT happening is \(P(not \ GG)=1-\frac{1}{16}=\frac{15}{16}\).
So your doubt was right: we do have 2 events (first pick and second pick) and the probability of each is indeed 5/20 (first green and second green), but the probability of both happening (both green) is the product (not the sum) of their individual probabilities.
Hope it's clear.
25 Aug 2010, 20:21
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Bunuel,
Thanks a ton for clarifying the doubt...
this question is from algebraworksheets....im sorry i cannot give you the url for the post....(printouts of those problems with explanation is the source)
one of the tutor has posted the ans and explanation
Thankyou ....once again....
Thanks a ton for clarifying the doubt...
this question is from algebraworksheets....im sorry i cannot give you the url for the post....(printouts of those problems with explanation is the source)
one of the tutor has posted the ans and explanation
Thankyou ....once again....
This Question is Locked Due to Poor Quality
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The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
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