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.
Number of problems , $n=50$,
Probability of not solving problem,

$q=1-p=1-\frac{4}{5}=\frac{1}{5}$,
Probability of solving problem, $p=\frac{4}{5}$

We have to find probability that he is unable to solve less then two problems

That is, $P(X<2)$

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

$={}^n{C}_0\left(p{)}^n\right(q{)}^0+{}^n{C}_2(p{)}^{n-1}.q^n$
$={}^{50}{C}_0{\left(\frac{1}{5}\right)}^0{\left(\frac{4}{5}\right)}^{50}+{}^{50}{C}_1{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^{49}$

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

$=\left(\frac{4}{5}{)}^{49}\right[\frac{4}{5}+\frac{50}{5}]$
$=\frac{54}{5}{\left(\frac{4}{5}\right)}^{49}$