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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!!

Debarghya Mukherjee
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2 Answers2

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how to calculate the other part

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.)

Graham Kemp
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See Probability distribution in the coupon collector's problem. This is the special case $m=3$. Substituting that into the general solution yields

$$ \def\stir#1#2{\left\{#1\atop#2\right\}} P(T=n)=\frac{3!}{3^n}\stir{n-1}{3-1}\;. $$

With $\stir n2=2^{n-1}-1$, this is

$$ P(T=n)=\frac{2^{n-1}-2}{3^{n-1}}\;. $$

This applies for $n\ge2$; clearly $P(T=1)=0$.

joriki
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