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