Question

A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is
(a) 64/64
(b) 49/64
(c) 40/64
(d) 24/64

Answer 1

Let A be the event of drawing are good article whereas B be the event of drawing are defected article

$\therefore \mathrm{P}\left(A\right)=\frac{10}{16}\mathrm{and}\mathrm{P}\left(B\right)=\frac{6}{16}$

Then P(A∪B) = P(A) + P(B) [∴ A and B are mutually exclusive]

$$\begin{gathered} \mathrm{i}.\mathrm{e}.\mathrm{P}\left(A\cup B\right)=\frac{10}{16}+\frac{6}{16} \\ \mathrm{P}\left(A\cup B\right)=\frac{16}{16}=1=\frac{64}{64} \end{gathered}$$

∴ Probability of one item being either good or has a defect = $\frac{64}{64}$

Hence, the correct answer is option A.