An urn contains $25$ balls of which $10$ balls are red and the remaining green. A ball is drawn at random from the urn, the colour is noted and the ball is replaced. If $6$ balls are drawn in this way, find the probability that:
(i) All the balls are red.
(ii) Not more than $2$ balls are green.
(iii) Number of red balls and green balls are equal.

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Solution

Ball is replaced each time

So probability of getting red ball in each draw $= \frac{10}{25} = \frac{2}{5}$

Probability of green ball in each draw $= \frac{15}{25} = \frac{3}{5}$

(i)Probablilty that all the balls are red $^{6} C_{6} {(\frac{2}{5})}^{6} = {(\frac{2}{5})}^{6} = \frac{64}{15625}$

(ii) Not more than $2$ green means no ball is green or $1$ ball if green or $2$ balls are green

Let required pronbability $= P (E)$

$P (E) =^{6} C_{0} {(\frac{2}{5})}^{6} +^{6} C_{1} (\frac{3}{5}) {(\frac{2}{5})}^{5} +^{6} C_{2} {(\frac{3}{5})}^{2} {(\frac{2}{5})}^{4} P (E) = {(\frac{2}{5})}^{4} {{(\frac{2}{5})}^{2} + 6 (\frac{3}{5}) (\frac{2}{5}) + 15 {(\frac{3}{5})}^{2}} P (E) = {(\frac{2}{5})}^{4} {\frac{4}{25} + \frac{36}{25} + \frac{135}{25}} P (E) = \frac{32}{625} \times 7 = \frac{224}{625}$

$(i i i)$ Number of red and green balls are equal , so we have $3$ red and $3$ green balls

So required probability $=$ $^{6} C_{3} \times {(\frac{2}{5})}^{3} \times^{6} C_{3} \times {(\frac{3}{5})}^{3} = 20 {(\frac{2}{5})}^{3} {(\frac{3}{5})}^{3} = \frac{864}{3125}$

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