Question

A bag contains $9$ discs of which $4$ are red, $3$ are blue and $2$ are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Find the probability that it is either red or blue.

• A. $\frac {2}{9}$
• B. $\frac {7}{9}$
• C. $\frac {1}{9}$
• D. $\frac {4}{9}$

Answer 1

Answer: (B)

There are $9$ discs in all so the total number of possible outcomes is $9$.
Let the events $A, B, C$ be defined as
$A :$ 'the disc drawn is red'
$B : $ 'the disc drawn is yellow'
$ C :$ 'the disc drawn is blue'
The number of red discs $= 4$, i.e., $n(A) = 4$
Hence $P(A) = \frac{4}{9}$
The number of blue discs $= 3$, i.e., $n(C) = 3$
Therefore, $P(C) = \frac{3}{9}$
$=\frac{1}{3}$
The event 'either red or blue' may be described by the set '$A$ or $C$'
Since, $A$ and $C$ are mutually exclusive events, we have
$P(A$ or $ C) = P(A \cup C) = P(A) + P(C)$
$ =\frac{4}{9} + \frac{1}{3} $
$= \frac{7}{9}$