A bag contains, $7$ different Black balls .and $10$ different Red balls, if one by one ball are randomely drawn untill all black balls are not drawn, then probability that this process is completed in $12 ^{th}$ draw, is equal to
$\frac{{^7{C_6}{\,^{10}}{C_6}}}{{^{17}{C_{12}}}} - \frac{{^1{C_1}}}{{^5{C_1}}}$
$\frac{{^7{C_6}{\,^{10}}{C_5}}}{{^{17}{C_{11}}}} - \frac{{^1{C_1}}}{{^6{C_1}}}$
$\frac{{^7{C_6}{\,^{10}}{C_10}}}{{^{17}{C_{11}}}} - \frac{{^1{C_1}}}{{^6{C_1}}}$
None
A bag contains $4$ white and $3$ red balls. Two draws of one ball each are made without replacement. Then the probability that both the balls are red is
Five numbers $x _{1}, x _{2}, x _{3}, x _{4}, x _{5}$ are randomly selected from the numbers $1,2,3, \ldots \ldots, 18$ and are arranged in the increasing order $\left( x _{1}< x _{2}< x _{3}< x _{4}< x _{5}\right)$. The probability that $x_{2}=7$ and $x_{4}=11$ is
If four vertices of a regular octagon are chosen at random, then the probability that the quadrilateral formed by them is a rectangle is
Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then $q- p$ is equal to
Four fair dice $D_1, D_2, D_3$ and $D_4$ each having six faces numbered $1,2,3,4,5$ and $6$ are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1, D_2$ and $D_3$ is