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 4/5, then the probability that he is unable to solve less than two problems is

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

Solution:

Total number of problems in the test= 50.

Probability of solving any problem = 4/5

and probability of not solving any problem = 1/5

Probability (a candidate is unable to solve less than two problems) =P (not solving zero problem)+ P (not solving one problem)

$=^{50} C_{0} {(\frac{4}{5})}^{50} +^{50} C_{1} (\frac{1}{5}) {(\frac{4}{5})}^{49} = {(\frac{4}{5})}^{49} [\frac{4}{5} + 10] = (\frac{54}{5}) {(\frac{4}{5})}^{49}$

Solution:

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