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 :

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

Answer 1

The correct option is B $\frac{54}{5}{\left(\frac{4}{5}\right)}^{49}$

Let $X$ denote the number of unsolved problems.
$n=50$,
Probability of not solving, $p=\frac{1}{5}$,
Probability of solving, $q=\frac{4}{5}$

We have to find $P(X<2)$

$P(X<2)=P(X=0)+P(X=1)$

$={}^{50}{C}_0{\left(\frac{1}{5}\right)}^0{\left(\frac{4}{5}\right)}^{50}+{}^{50}{C}_1{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^{49}$

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

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