An urn contains 10 black 10 white and 10 blue balls . Balls are drawn randomly with replacement Let T be the minimum number of draws required to get balls of all different colors .Find distribution of T
P(T=n)=P(T>=n)-P(T>n)
now P(T>n)=Number of ways we get only 2 balls of different colour -Number of ways we get 1 ball of different colour in n draws
only balls of two different colors are obtained with probability $$\sum_{x=0}^n{n\choose x}(\frac {1}{3})^x(\frac {1}{3})^{n-x} + \sum_{x=0}^n {n\choose x}(\frac {1}{3})^x(\frac {1}{3})^{n-x} + \sum_{x=0}^n{n\choose x}(\frac {1}{3})^x(\frac {1}{3})^{n-x}$$
-only black or blue or white balls are obtained with probability $(\frac {1}{3})^n +(\frac {1}{3})^n +(\frac {1}{3})^n$
how to calculate the other part i.e. P(T>=n) please help!!
You do not need to.
The event $\{T\,{=}\,n\}$ is the event of drawing one from three colours in the last draw, and only the remaining two colors among the preceding $n\,{-}\,1$ draws but those must not be all of the same colour. (Clearly, $n$ should be at least $3$).
You evaluate the probability for this using the principle of inclusion and exclusion.
(There is no need to involve series in this.)