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*Answer whether the following statements are True or False.*

There are $N$ balls in a box, out of which $n$ are blue $(1 < n < N)$ and the rest are red. Balls are drawn from the box one by one at random, and discarded. Then the probability of picking all the blue balls in the first $n$ drawn is the same as the probability of picking all the red balls in the first $(N – n)$ draws.

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*Total balls=*N

*blue balls=* n

*red balls = *N-n

*probability of n blue balls in first n draw *

* $\frac{^{n}C_{n}}{^{N}C_{n}}{}$*

*probailty of (N-n) red balls in first (N-n) draw*

$\frac{^{(N-n)}C_{(N-n)}}{^{N}C_{(N-n)}}$

according to question

*$\frac{^{n}C_{n}}{^{N}C_{n}}{}$=*$\frac{^{(N-n)}C_{(N-n)}}{^{N}C_{(N-n)}}$

$\implies\frac{1}{^{N}C_{(n)}}=\frac{1}{^{N}C_{(N-n)}}$

$\implies\frac{(N-n)! n!}{N!}=\frac{(N-(N-n))!(N-n)!}{N!}$

$\implies\frac{(N-n)!n!}{N!}=\frac{(N-n)!n!}{N!}$

LHS=RHS

*Second method*

*if total balls =5*

*blue balls =2*

*then red balls = 5-2=3*

*probabilty of picking all blue balls in first 2 draw =*$\frac{^{2}C_{2}}{^{5}C_{2}}$

*probabilty of picking all red balls in first 3 draw =* $\frac{^{3}C_{3}}{^{5}C_{3}}$

*according to question*

$\implies\frac{^{2}C_{2}}{^{5}C_{2}}=\frac{^{3}C_{3}}{^{5}C_{3}}$

$\implies\frac{1}{10}=\frac{1}{10}$

*LHS=RHS*