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.
$$\frac{1}{5}$$
B.
$$\frac{5}{{24}}$$
C.
$$\frac{3}{{10}}$$
D.
$$\frac{1}{{40}}$$
Answer :
$$\frac{3}{{10}}$$
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 is
$$ = \frac{{{}^6{C_1} \times {}^4{C_2}}}{{{}^{10}{C_3}}} = \frac{{6 \times 6 \times 6}}{{10 \times 9 \times 8}} = \frac{3}{{10}}$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$