For an initial screening of an entrance exam, a candidate is given fifty problems to solve. If the…
For an initial screening of an entrance exam, 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
$\frac{201}{5}\left(\frac{1}{5}\right)^{49}$
$\frac{316}{25}\left(\frac{4}{5}\right)^{48}$
$\frac{54}{5}\left(\frac{4}{5}\right)^{49}$
$\frac{164}{25}\left(\frac{1}{5}\right)^{48}$
Solution
$q=$ Probability that the candidate can solve any problem $=\frac{4}{5}$
$\mathrm{p}=1-\frac{4}{5}=\frac{1}{5}$
Also, $\mathrm{n}=50$
$\begin{aligned} \therefore \quad & \text { Required probability }=\mathrm{P}(\mathrm{X} < 2) \\ & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ & ={ }^{50} \mathrm{C}_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^{50}+{ }^{50} \mathrm{C}_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^{49} \\ & =\left(\frac{4}{5}\right)^{50}+50\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^{49} \\ & =\left(\frac{4}{5}+\frac{50}{5}\right)\left(\frac{4}{5}\right)^{49} \\ & =\left(\frac{54}{5}\right)\left(\frac{4}{5}\right)^{49}\end{aligned}$