A bag contains 19 red balls and 19 black ball | vedclass

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(A) Let $R$ be the number of red balls and $B$ be the number of black balls. Initially,$R = 19$ and $B = 19$.In each step,two balls are removed:$1$. I

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $R$ be the number of red balls and $B$ be the number of black balls. Initially,$R = 19$ and $B = 19$.In each step,two balls are removed:$1$. If two red balls are removed,$R$ decreases by $2$ $(R 	o R-2, B 	o B)$.$2$. If two black balls are removed,$B$ decreases by $2$ $(R 	o R, B 	o B-2)$.$3$. If one red and one black ball are removed,the black ball is discarded and the red ball is returned $(R 	o R, B 	o B-1)$.Notice that in all cases,the number of black balls $B$ decreases by either $0$ or $2$ or $1$. Specifically,the parity of the number of black balls changes if we remove one black ball,but the process continues until no more pairs can be formed.Crucially,the number of red balls $R$ only changes parity if we remove two red balls. Since we start with $19$ red balls (odd) and $19$ black balls (odd),and the process must terminate,we observe that the number of black balls will eventually reach $0$. Since we cannot remove a red ball without another red ball,and we only discard red balls in pairs,the number of red balls will remain odd. Thus,the process must terminate with $1$ red ball remaining. Therefore,the probability is $1$.

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