JEE Mains · Maths · STD 12 - 13. probability
A bag contains \(30\) white balls and \(10\) red balls. \(16\) balls are drawn one by one randomly from the bag with replacement. If \(X\) be the number of white balls drawn, then \(\left( {\frac{{{\rm{mean\, of\, X}}}}{{{\rm{standard\, deviation\, of\, X}}}}} \right)\) is equal to
- A \(4\)
- B \(4\sqrt 3 \)
- C \(3\sqrt 2 \)
- D \(\frac{{4\sqrt 3 }}{3}\)
Answer & Solution
Correct Answer
(B) \(4\sqrt 3 \)
Step-by-step Solution
Detailed explanation
There are 30 white balls and 10 red balls \(P(\text { white ball })=\frac{30}{40}=\frac{3}{4}=p\) \(\Rightarrow \mathrm{q}=\frac{1}{4}\) \(\frac{\operatorname{mean}(\mathrm{x})}{\text { standard deviation }(\mathrm{x})}=\frac{\mathrm{np}}{\sqrt{\mathrm{npq}}}\)…
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