Question

A bag contains 12 white and black balls out of which $x$ are black. If one ball is drawn at random, what is the probability it will be a white ball?

• A. $\frac{\left(x\right)}{12}$
• B. $\frac{(12+x)}{12}$
• C. $\frac{(12-x)}{12}$
• D. $\frac{\left(x\right)}{2}$

Answer 1

The correct option is C $\frac{(12-x)}{12}$

Given, total number of balls = 12

Number of black balls $=x$

We know, for an event E,
$P\left(E\right)=\frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$

Then, P(getting a black ball) $=\frac{x}{12}$

We know that the sum of probabilities of all possible outcomes of an event is 1.

So, probability of getting a white ball
$=1-\frac{x}{12}$
$=\frac{(12-x)}{12}$