Question

A bag contains 6 red, 8 black and 10 white balls. One ball is drawn at random. Find the probability that the ball drawn is not red.
1. \(\frac14\)
2. \(\frac34\)
3. \(\frac23\)
4. \(\frac{1}{12}\)

Answer 1

Correct Answer - Option 2 : \(\frac34\)

Concept:

In a random experiment, let S be the sample space and let E ⊆ S .Then, E is an event.

The probability of occurrence of E is defined as, \(\rm P(E)=\frac{n(E)}{n(S)}\), where, n(E) = number of elements in E and n(S) = number of possible outcomes.

P(Getting E) = 1 - P(Not getting E) OR P(Event happenning ) = 1 - P(Event not happenning) 

 

Calculation:

Total number of balls = (6 + 8 + 10) = 24.

Let S be the sample space. Then, n(S) = 24

Let E be the event of drawing a red ball. Then, n(E) = 6

P(getting a red ball) = \(\rm P(E)=\frac{n(E)}{n(S)}\)

\(=\frac{6}{24}=\frac{1}{4}\)

Now, probability of not getting red ball = 1 - P(getting a red ball)

\(= \rm 1-\frac{1}{4}\\ =\frac{3}{4}\)

Hence, option (2) is correct.