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?

Question

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? (A) 1/5 (B) 5/24 (C) 3/10 (D) 1/40

Tardigrade Admin · Accepted Answer

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 _{2}} = \frac{6 \times 6 \times 6}{10 \times 9 \times 8} = \frac{3}{10}

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?

1/5

5/24

3/10

1/40

Total number of selecting 3 components out of 10=10C3​. Out of 3 selected components two defective pieces can be selected in 4C2​ ways and one non-defective piece will be selected in 6C1​ ways, hence, Required probability =10C2​6C1​×4C2​​=10×9×86×6×6​=103​