For an entry to a certain course, a candidate is given twenty problems to solve. If the probability that the candidate can solve any problem is \(\frac{3}{7}\), then the probability that he is unable to solve at most two problem is

\(\mathrm{q}=\) probability that the candidate can solve the problem \(=\frac{3}{7}\)
\(\therefore \quad \mathrm{p}=1-\frac{3}{7}=\frac{4}{7}\)
Also, \(\mathrm{n}=20\).
\(\therefore \text { Required probability } \)
\( =\mathrm{P}(\mathrm{X} \leq 2) \)
\( =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \)
\( ={ }^{20} \mathrm{C}_0 \mathrm{p}^{20} \mathrm{q}^0+{ }^{20} \mathrm{C}_1 \mathrm{p}^{19} \mathrm{q}+{ }^{20} \mathrm{C}_2 \mathrm{p}^{18} \mathrm{q}^2 \)
\( =1 \times\left(\frac{4}{7}\right)^{20} \times\left(\frac{3}{7}\right)^0+20 \times\left(\frac{4}{7}\right)^{19} \times\left(\frac{3}{7}\right)\) \(+~190\left(\frac{4}{7}\right)^{18} \times\left(\frac{3}{7}\right)^2 \)
\( =\left(\frac{4}{7}\right)^{18}\left[\frac{16}{49}+\frac{240}{49}+\frac{1710}{49}\right] \)
\( =\left(\frac{4}{7}\right)^{18} \cdot \frac{1966}{49}\)