A) 1/5
B) 5/24
C) 3/10
D) 1/40
Correct Answer: C
Solution :
[c] Total number of selecting 3 components out of \[10{{=}^{10}}{{C}_{3}}.\] Out of 3 selected components two defective pieces can be selected in \[^{4}{{C}_{2}}\] ways and one non-defective piece will be selected in \[^{6}{{C}_{1}}\]ways. Hence, required probability \[=\frac{^{6}{{C}_{1}}{{\times }^{4}}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{6\times 6\times 6}{10\times 9\times 8}=\frac{3}{10}\]
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