- ✓$1$
- B$2$
- C$3$
- D$4$
$=\sin \left(\sin ^{-1}\left(\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}}\right)\right)$
$=\left(\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos ^2 \theta-\sin ^2 \theta}}\right)$
$=\left(\frac{\sin \theta}{\sqrt{\cos ^2 \theta}}\right)$
$=\left(\frac{\sin \theta}{\cos \theta}\right)=\tan \theta$
$\text { Now, } \frac{d}{d(\tan \theta)}(\tan \theta)=1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(1)$ Probability that the selected bag is $B _3$ and the chosen ball is green equals $\frac{3}{10}$
$(2)$ Probability that the chosen ball is green equals $\frac{39}{80}$
$(3)$ Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$
$(4)$ Probability that the selected bag is $B_3$, given that the chosen balls is green, equals $\frac{5}{13}$