A bag contains 7 different black balls and 10 | vedclass

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Question

(A) To complete the process in the $12^{th}$ draw,the $12^{th}$ ball drawn must be the $7^{th}$ black ball. This implies that in the first $11$ draws,

Vedclass · Accepted Answer

$\frac{^7C_6 	imes ^{10}C_5}{^{17}C_{11}} 	imes \frac{^1C_1}{^6C_1}$

Vedclass · Answer

(A) To complete the process in the $12^{th}$ draw,the $12^{th}$ ball drawn must be the $7^{th}$ black ball. This implies that in the first $11$ draws,exactly $6$ black balls and $5$ red balls must have been drawn. The probability of drawing $6$ black balls and $5$ red balls in the first $11$ draws is given by $\frac{^7C_6 	imes ^{10}C_5}{^{17}C_{11}}$. After $11$ draws,$1$ black ball and $5$ red balls remain in the bag (total $6$ balls). The probability that the $12^{th}$ draw results in the last black ball is $\frac{^1C_1}{^6C_1}$. Thus,the total probability is $\frac{^7C_6 	imes ^{10}C_5}{^{17}C_{11}} 	imes \frac{^1C_1}{^6C_1}$.

$A$ bag contains $7$ different black balls and $10$ different red balls. If balls are drawn one by one randomly until all black balls are drawn,what is the probability that this process is completed in the $12^{th}$ draw?

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