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??

pass

Look at your FE book, I think you are going to attend in that exam.

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.

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.

I know it is greater than 0.8. But i couldn't figure out the exact probability.

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.

my bad, double entry.

Your solution didn't convince me parbatya. But I  appreciate your thought. Could you explain me how the probability in second is .125?

Sorry, That should be 0.1 and total probabilty 0.9

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

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..

Thanks parbatya and shivanagar. The probability is 90%. I got it.

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

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 !!!

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.

oh man i have this test on monday

 

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

or it could be:
if you miss out in the first opportunity, you will lose your BALLS? HAHAHAHAHAHAHA

I hope that someone will post the analogy of this question very soon. It is related to the present hot issue. Guess.