# Engineering company has a t ofchecking compressive strength fo

Engineering company has a t of checking compressive strength for 100 concrete cubes. The results revealed that 85 cubes passed the compressive strength test successfully and 15 cubes failed in the test. If 10 cubes are selected at random to be inspected by the company, determine the probability that the 8 cubes will pass the test and 2 cubes will fail in the test by using the Combinatorial Analysis. *0.4522
a) 0.3415
b) 0.6553
c) 0,1156
0.2919

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James Obrien

Total number of cubes=100
Number of cubes passed the test successfully=85
Number of 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 $$\displaystyle={\frac{{{F}{a}{v}{{or}}{a}{b}le\ {c}{a}{s}{e}{s}}}{{{T}{o}{t}{a}{l}\ {p}{o}{s}{s}{i}{b}le\ {c}{a}{s}{e}{s}}}}$$
$$\displaystyle={\frac{{^{\left\lbrace{85}\right\rbrace}{C}_{{8}}\times^{{15}}{C}_{{2}}}}{{^{\left\lbrace{100}\right\rbrace}{C}_{{{10}}}}}}$$
$$\displaystyle{\frac{{{4.8125}\times{10}^{{{10}}}\times{105}}}{{{1.731}\times{10}^{{{13}}}}}}$$
$$\displaystyle={\frac{{{2.7801}\times{105}}}{{{1000}}}}$$
$$\displaystyle={\frac{{{291.92}}}{{{1000}}}}$$
$$\displaystyle={0.2919}$$
Thus, the required probability that 8 cubes will pass the test and 2 cubes will
$$\displaystyle{f}{a}{i}{l}\ in\ {t}{h}{e\ }{t}{e}{s}{t}\ {i}{s}\ {0.2919}.$$
Hence,option(e) is correct.