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  4. 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?

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Q. 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?

Probability

Solution:

Total number of selecting 3 components out of $10 = {^{10}C_3}$.
Out of 3 selected components two defective pieces can be selected in $^4C_2$
ways and one non-defective piece will be selected in $^6C_1$ ways, hence,
Required probability
$ = \frac{^{6}C_{1} \times^{4}C_{2}}{^{10}C_{2}} = \frac{ 6 \times 6\times 6}{10\times 9\times 8} = \frac{3}{10}$