A box contains 10 identical electronic components of which 4 are defective. If 3 components are selected at random from the box in succession, without replacing the units already drawn, what is the probability that two of the selected components are defective ?

Options

A. `1//5`
B. `5//24`
C. `3//10`
D. `1//40`

Answer

C

Explanation

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 `= (""^(6)C_(1) xx ""^(4)C_(2))/(""^(10)C_(3)) = (6 xx 6 xx 6)/(10 xx 9 xx 8) = 3/10`