MCQ
If ${\sin ^{ - 1}}\frac{x}{5} + {\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2},$ then $x = $
- A$4$
- B$5$
- C$1$
- ✓$3$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow \frac{x}{5}=\sin \left(\cos ^{-1} \frac{4}{5}\right)$
$\Rightarrow \frac{x}{5}=\sin \left(\sin ^{-1} \frac{3}{5}\right)$
$\Rightarrow \frac{x}{5}=\frac{3}{5}$
$\Rightarrow x=3$
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$190$ persons had symptom of fever,
$220$ persons had symptom of cough,
$220$ persons had symptom of breathing problem,
$330$ persons had symptom of fever or cough or both,
$350$ persons had symptom of cough or breathing problem or both,
$340$ persons had symptom of fever or breathing problem or both,
$30$ persons had all three symptoms (fever, cough and breathing problem).
If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is. . . . .