Probability Question 366
Question: A box contains 10 identical electronic components of which 4 are defective. If 3 components are selected at random form 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
Show Answer
Answer:
Correct Answer: C
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 $ ^{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} $
BETA
