# An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability for the following: (i) all will bear 'X' mark. (ii) not more than 2 will bear 'Y' mark. (iii) at least one ball will bear 'Y' mark. (iv) the number of balls with 'X' mark and 'Y' mark will be equal.

Toolbox:
• A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
Given an urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'.
P (drawing an X) $= p = \large\frac{10}{25} = \frac{2}{5} \rightarrow$$q = 1- p = \large\frac{3}{5} Given that 6 balls are to be drawn at random, it is a case of Bernoulli trials (n=6), with p(X=r)=C^{6}_{r} \large(\frac{2}{5})^{r} (\frac{3}{5})^{6-r} (i) all will bear the X mark P (X = 6) = C^{6}_{6} \large(\frac{2}{5})^{6} (\frac{3}{5})^{6-6} = (\frac{2}{5})^6 (ii) not more than 2 balls will bear the Y mark. P (not more than 2 balls with Y mark = not less than 4 bear X mark) = P (X \geq 4) = P (X=4) + P (X=5) + P(X=6) =$$\large\;c^6_4\;(\frac{2}{5})^4\;(\frac{3}{5})^2+c^6_4\;(\frac{2}{5})^5\;(\frac{3}{5})^1\;+\;c^6_6\;(\frac{2}{5})^6\;(\frac{3}{5})^0$$ =$$\;15\;\large \frac{2^4x3^2}{5^6}$$$$+\;6\;\large\frac{2^5 \times 3}{5^6}+\frac{2^6}{5^6}$$ =$$\;7\large (\frac{2}{5})^4$$ (iii) at least 1 ball will bear Y mark. P (at least 1 Y mark) = P (at most 5 X marks) = 1 - P (X = 6) =$$\;1-\large c^6_6\;(\frac{2}{5})^6\;(\frac{3}{5})^0$$ =$$\;1-\large( \frac{2}{5})^6$$ (iv) number of balls w/ X mark and Y mark will be equal. P ( same number of X and Y out of 6) = P (exactly 3 successes and 3 failures) = P (3) P (X = 3) = C^{6}_{3} \large(\frac{2}{5})^{3} (\frac{3}{5})^{6-3}$$= 20 \large (\frac{2}{5})^3 (\frac{3}{5})^3 = \frac{864}{3125}$