Question

A box contains 10 good articles and 6 with defects. One item is draqn at random. The Probability that it is either good of has a defect is

• A. $\frac{64}{64}$
• B. $\frac{49}{64}$
• C. $\frac{40}{64}$
• D. $\frac{24}{64}$

Answer 1

The correct option is A

$\frac{64}{64}$

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

Here, $P\left(A\right)=\frac{10}{10+6}=\frac{10}{16}$

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

The events A and B are mutually exclusive. Thus, the required probability is $P(A\cup B)=P\left(A\right)+P\left(B\right)$

$\Rightarrow P(A\cup B)=\frac{10}{16}+\frac{6}{16}=\frac{16}{16}=1$

Hence, the correct option is (a)