Engineering company has a task of checking compressive strength for 100 concrete cubes. The results...

Total number of cubes ${\displaystyle =100}$
Number of cubes passed the test successfully $=85$

Number of cubes failed in the test $=15$
It is given that if 10 cubes are selected at random to be inspected by the company then we have to find the probability that 8 cubes will pass the test and 2 cubes will fail in the test.
Total possible cases ${\displaystyle {=}^{100}{C}_{10}}$
Favorable cases ${\displaystyle {=}^{85}{C}_8{\times }^{15}{C}_2}$
Required probability $=\frac{Favorablecases}{Totalpossiblecases}$
$=\frac{{}^{85}{C}_8{\times }^{15}{C}_2}{{}^{100}{C}_{10}}$
${\displaystyle =\frac{4.8125\times {10}^{10}\times 105}{1.731\times {10}^{13}}}$
${\displaystyle =\frac{2.7801\times 105}{1000}}$
${\displaystyle =\frac{291.92}{1000}}$
${\displaystyle =0.2919}$
Thus, the required probability that 8 cubes will pass the test and 2 cubes will fail in the test is 0.2919.
Hence, option (e) is correct