MCQ
Suppose four balls labelled $1,2,3,4$ are randomly placed in boxes $B_1, B_2, B_3, B_4$. The probability that exactly one box is empty is
- A$\frac{8}{256}$
- ✓$\frac{9}{16}$
- C$\frac{27}{256}$
- D$\frac{9}{64}$
Four balls $1,2,3,4$ are randomly placed in boxes $B_1, B_2, B_3, B_4$.
Probability of exactly one box is empty is
$\frac{{ }^4 C_1 \times \frac{4 !}{1 ! \times 2 !} \times \frac{1}{2 !} \times 3 !}{4^4}=\frac{4 \times 6 \times 6}{4 \times 4 \times 4 \times 4}=\frac{9}{16}$
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The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:
$\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
$\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$
$\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
$\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$