Assume that we have 2 boxes each contains 5 balls , there is one ball among all these ball that is different from other balls , we want to choose 3 ball randomly from each box ( we will put the ball back after checking whether its the one or not ) , what's the probability of finding the ball ?
My answer : the different ball can be in either box one or two with P of 1/2 we choose 3 from each box randomly so the probability for each ball to be chosen is 1/5 so the P of finding the ball in first box = 1/2 (1/5 )^3 same for the second box so the P of finding the ball in two boxes is = (1/5)^3 + (1/5)^3 but if we decide to choose 4 ball instead of 3 ball the chance of finding the ball seems to be more ? but with this approach its going down ! 2*(1/5)^4 , so guess there must be something wrong with my way , or is it because of randomness that produces many other mutations of ball being chosen ? ( for example same ball being chosen each time ...)
UPDATE : i guess it might be possion hmm ?
In one of the boxes you will find nothing for sure. The probability that you don't find the special ball in the box containing it is $\left({4\over5}\right)^3={64\over125}$. Therefore the probability that you have found the special ball after at most six trials is ${61\over125}=0.488$.