JEE Mains · Maths · STD 12 - 13. probability
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is \(\frac{4}{5}\) , then the probability that he is unable to solve less than two problems is
- A \(\frac{{164}}{{25}}{\left( {\frac{1}{5}} \right)^{48}}\)
- B \(\frac{{201}}{{5}}{\left( {\frac{1}{5}} \right)^{49}}\)
- C \(\frac{{54}}{{5}}{\left( {\frac{4}{5}} \right)^{49}}\)
- D \(\frac{{316}}{{25}}{\left( {\frac{4}{5}} \right)^{48}}\)
Answer & Solution
Correct Answer
(C) \(\frac{{54}}{{5}}{\left( {\frac{4}{5}} \right)^{49}}\)
Step-by-step Solution
Detailed explanation
Total problems \(=50\) \(\mathrm{P}(\text { Solving })=\frac{4}{5}\) \(P(\text { Not solving })=\frac{1}{5}\) \(\mathrm{P}\) (unable to solve less than two problems) \(=\mathrm{P}\) (not solving one problem) \(+\mathrm{P}\) (not solving zero problem)…
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