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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.

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

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