For an initial screening of an admission test, a candidate is giv… - Vidyadip

MCQ

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
$\frac{{201}}{{5}}{\left( {\frac{1}{5}} \right)^{49}}$
✓
$\frac{{54}}{{5}}{\left( {\frac{4}{5}} \right)^{49}}$
D
$\frac{{316}}{{25}}{\left( {\frac{4}{5}} \right)^{48}}$

✓

Answer

Correct option: C.

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

c
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)

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

$=\frac{4^{50}}{5^{50}}+50 \cdot \frac{4^{40}}{5 \cdot 5^{49}}$

$=\left(\frac{4}{5}\right)^{50}+10 \cdot\left(\frac{4}{5}\right)^{49}$

$=\left(\frac{4}{5}\right)^{49}\left(\frac{4}{5}+10\right)$

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

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