For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate solve any problem is \frac{4}{5}, then the probability that he is unable to solve less than two problems is :

Engineering

Mathematics

Binomial Probability Theorem

Question

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate solve any problem is $\frac{4}{5}$ , then the probability that he is unable to solve less than two problems is :

$\frac{201}{5} {(\frac{1}{5})}^{49}$

$\frac{54}{5} {(\frac{4}{5})}^{49}$

$\frac{316}{25} {(\frac{4}{5})}^{48}$

$\frac{164}{25} {(\frac{1}{5})}^{48}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Total problems = 50

P(Solving) = $\frac{4}{5}$

P(Not solving) = $\frac{1}{5}$

P(unable to solve less than two problems)

= P(not solving one problem) + P(not solving zero problem)

$=^{50} C_{0} {(\frac{1}{5})}^{0} {(\frac{4}{5})}^{50} +^{50} C_{1} {(\frac{1}{5})}^{1} {(\frac{4}{5})}^{49}$

$= \frac{4^{50}}{5^{50}} + 50 . \frac{4^{49}}{5 {.5}^{49}}$

$= {(\frac{4}{5})}^{50} + 10 . (\frac{4}{5} + 10)$

$= \frac{54}{5} . {(\frac{4}{5})}^{49}$