Question

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. $\frac{201}{5}{\left(\frac{1}{5}\right)}^{49}$
• B. $\frac{164}{25}{\left(\frac{1}{5}\right)}^{48}$
• C. $\frac{316}{25}{\left(\frac{4}{5}\right)}^{48}$
• D. $\frac{54}{5}{\left(\frac{4}{5}\right)}^{49}$

Answer 1

Answer: (D)

Let, $X=$ The number of questions, student are unable to solve.
$p=P$ (the student answers a question correctly) $=\frac{4}{5}$
$\Rightarrow q=1-p=\frac{1}{5}$ (probability that the student does not answer a question correctly)
Also, $n=50$ (Total number of question)
Hence, required probability:
$P(X<2)$
$=P\left(X=0\right)+P\left(X=1\right)$
$={}^{50}{C}_0{\left(\frac{4}{5}\right)}^{50}+{}^{50}{C}_1{\left(\frac{4}{5}\right)}^{49}\left(\frac{1}{5}\right)$
$=\frac{54}{5}\left(\frac{4}{5}\right)$