Question

A bag contains 3 red, 4 blue and 3 green balls. A ball is drawn at random. Event E represents `not drawing a blue ball'. Then find the probability of \(\bar{E}\) (complementary event of E).

• A. \(\dfrac{3}{5}\)
• B. \(\dfrac{3}{10}\)
• C. \(\dfrac{7}{10}\)
• D. \(\dfrac{2}{5}\)

Hint:

The key here is reading carefully: the question asks for \(P(\bar{E})\) (the complement), not \(P(E)\). Event \(E\) = ``not blue'' \(\Rightarrow\) \(\bar{E}\) = ``blue''. The complementary probability law \(P(\bar{E}) = 1 - P(E)\) is always the fastest path. Make sure you identify what \(E\) is and what \(\bar{E}\) is before computing.

Answer 1

The Correct Option is D

Concept:
The

complementary event \(\bar{E}\) of an event \(E\) is the event that \(E\) does not occur. The fundamental relation is:
Alternatively, if \(E = \text{``not drawing a blue ball''}\), then \(\bar{E} = \text{``drawing a blue ball''}\).

Step 1: Count the total number of balls.

• Red balls: 3.

• Blue balls: 4.

• Green balls: 3.

• Total balls: \(3 + 4 + 3 = 10\).

Step 2: Identify event \(E\) and its complement \(\bar{E}\).

• Event \(E\): ``not drawing a blue ball'' (i.e.
drawing a red or green ball).

• Complement \(\bar{E}\): ``drawing a blue ball''.

Step 3: Compute \(P(E)\).
Favourable outcomes for \(E\) (not blue): red or green \(= 3 + 3 = 6\) balls.

Step 4: Compute \(P(\bar{E})\) using the complementary rule.


Alternative Direct Method:
Favourable outcomes for \(\bar{E}\) (drawing a blue ball) = 4 blue balls.
Both methods confirm the same answer.

Step 5: Check all options.

• \(\dfrac{3}{5} = P(E)\), not \(P(\bar{E})\).

• \(\dfrac{3}{10}\): This would be \(P(\text{red only})\) or \(P(\text{green only})\).

• \(\dfrac{7}{10}\): No direct interpretation that fits.

• \(\dfrac{2}{5} = \dfrac{4}{10} = P(\bar{E})\).