[VIEWED 4453
TIMES]
|
SAVE! for ease of future access.
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 2:16
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
The defintion of sucess is to select the red ball. There are 8 red balls and 2 yellow balls in first container and 5 red balls and 5 yellow balls in second container. The ball will be selected from the first container first. If red ball is selected second trial will not be made. If not second trial will be done from second container. In such case what is the probability of sucess?
Do you guys have any guess to the analogy of this probability question??
|
|
|
|
marconi
Please log in to subscribe to marconi's postings.
Posted on 04-04-08 3:50
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
|
|
|
parbatya
Please log in to subscribe to parbatya's postings.
Posted on 04-04-08 3:52
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Look at your FE book, I think you are going to attend in that exam.
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 4:22
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
parbatya i am expecting the solution of this problem. Please do not refer me to FE book. By the way i already passed the FE exam.
|
|
|
parbatya
Please log in to subscribe to parbatya's postings.
Posted on 04-04-08 4:37
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
I guess, probability of picking red ball is always 0.8 . If you go to next container you will get less value, so the higher value is probability of success.
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 4:51
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
I know it is greater than 0.8. But i couldn't figure out the exact probability.
|
|
|
parbatya
Please log in to subscribe to parbatya's postings.
Posted on 04-04-08 4:57
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Then it is 0.925
Probabilty of getting right in one container = 0.8
If it fails, probability of getting red on next container is = 2/10 x 5/10 = .125
The two will be added as both cases are likely to occur. It is like adding up two prob values when drawing the same colour ball from two different containers.
Just a thought. Mathematician can explain more.![](http://www.sajha.com/sajha/chat/smileys/m-biggrin.gif)
|
|
|
parbatya
Please log in to subscribe to parbatya's postings.
Posted on 04-04-08 5:01
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
my bad, double entry.
Last edited: 04-Apr-08 05:01 PM
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 5:06
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Your solution didn't convince me parbatya. But I appreciate your thought. Could you explain me how the probability in second is .125?
|
|
|
parbatya
Please log in to subscribe to parbatya's postings.
Posted on 04-04-08 5:08
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Sorry, That should be 0.1 and total probabilty 0.9
|
|
|
shivanagar
Please log in to subscribe to shivanagar's postings.
Posted on 04-04-08 5:16
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
I guess it would be 0.9 0.8+ 0.2 *0.5=0.9 and to validate the results, I write a code in R, (you can do it in EXCEL to, to randomize events and see output, with 1000 samples each time, i did 10 times. and in those 10 times I got 0.9,0.89,0.91, 0.8667, 0.893, 0.9, 0.91, 0.8667, 0.92, 0.87
|
|
|
lootekukur
Please log in to subscribe to lootekukur's postings.
Posted on 04-04-08 5:20
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Here you go: probability of success in first attempt, P(s1)=8/10
probability of failure in first attempt, P(f1)=2/10
probability of success in second attempt, P(s2)=5/10
probability of failure in second attempt, P(f2)=5/10
The user goes for a second attempt only if he fails at the first. You
can use the conditional probability of success at second attempt given
the first is a failure, P(s2|f1)=P(s2^f1)/P(f1)= P(s2)*P(f1)/p(f1) = p(s2)=5/10.
(ie. same as P(s2)...other way of looking at this is because the second
success probability is INDEPENDENT of the failure at the first)
But the user will go to the second attempt ONLY when he fails at the
first. Therefore success at the second attempt after failure from the
first will be, P(s2|f1 ^f1)= P(s2|f1)*P(f1)=5/10*2/10=0.1
Hence the total success probability will be P(s)=P(s1+ P(s2|f1 ^f1)=8/10+0.1 = 0.9
(Note: ^ means AND operator)
However, be informed that it's been EONS I haven 't played with probabilities...so a double check wouldn't hurt.
Hope it helps..
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 5:22
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Thanks parbatya and shivanagar. The probability is 90%. I got it.
|
|
|
baglung_in_mass
Please log in to subscribe to baglung_in_mass's postings.
Posted on 04-04-08 5:24
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Define the following events: S : Success R1: Selecting a red ball from first container R2: Selecting a red ball from second container R0: NOT selecting a red ball from first container Symbols: * is intersection + is union And note that R0 and R2 are independent events, i.e., P(R0.R2) = P(R0)P(R2) Now, P(S) = P{R1+ (R0.R2)} = P(R1) + P(R0).P(R2) = (8/10) + [(2/10)(5/10)] = 0.9
|
|
|
sajhafan
Please log in to subscribe to sajhafan's postings.
Posted on 04-04-08 5:33
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
It depends upon what type container you are using.............if its a ballot container and its in saptari then your probability of success is very slim.......I'm afraid you will see neither red balls nor yellow balls. So please verify the question............I'm confused !!!
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 5:48
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Thanks lootekukur and baglung in mass. But actually nobody tried to catch the analogy portion which i had asked later. Sajhafan tried but that is not the right one.
|
|
|
gundaa
Please log in to subscribe to gundaa's postings.
Posted on 04-04-08 5:52
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
oh man i have this test on monday
|
|
|
lootekukur
Please log in to subscribe to lootekukur's postings.
Posted on 04-04-08 5:58
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Anilbro, Analogy? I got the morale : If you have more balls in your favor, the probability of you succeeding is high? More the balls, all the more better? HAHAHAHA
|
|
|
lootekukur
Please log in to subscribe to lootekukur's postings.
Posted on 04-04-08 6:01
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
or it could be: if you miss out in the first opportunity, you will lose your BALLS? HAHAHAHAHAHAHA
|
|
|
anilbro
Please log in to subscribe to anilbro's postings.
Posted on 04-04-08 6:01
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
I hope that someone will post the analogy of this question very soon. It is related to the present hot issue. Guess.
|
|