- A$\frac{1}{3} \,xyz (x + y) (y + z) (z + x)$
- B$\frac{1}{4} \,xyz (x + y - z) (y + z - x)$
- ✓$\frac{1}{12} \, xyz (x - y) (y - z) (z - x)$
- Dnone
$R_1 \rightarrow R_1 - R_2 \, and \, R_2 \rightarrow R_2 - R_3$
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If
$\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ and $\text{P}(\text{B})=\frac{17}{20},$ then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ equas:$\frac{14}{17}$
$\frac{17}{20}$
$\frac{7}{8}$
$\frac{1}{8}$
$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$, the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from bo $B _2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Give the answer question $1$ and $2.$