Question

A bag contains 9 black and 12 white balls.One ball is drawn at random.What is the probability that the ball drawn is black

Answer 1

hi dear here is the answer

total balls =9+12
= 21

p(black ball) =9/21
= 3/7

p( white Ball) = 12/21
=4/7


verification
p(E)+p(not E) =1

p(black balls)+p(white balls) = 3/7+4/7
=7/7 =1

hence proved

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Answer 2

Answer:

The Correct Answer would be $\frac{3}{7}$ . If a bag contains 9 black and 12 white balls and one ball is drawn at random, the probability of it being black is $\frac{3}{7}$.

Step-by-step explanation:

Given: Number of black balls in the bag = 9

           Number of white balls in the bag = 12

To Find: Probability of drawing a black ball.

Now, the bag contains 9 black and 12 white balls. Therefore, total number of balls present in the bag = 9 + 12 = 21.

Now we have to find the probability of ball drawn to be black.

From the formula of probability, we know:

        Probability of favorable outcome = Number of favorable                                                     outcomes/Total number of outcomes

Let the probability of black ball drawn be denoted by P(B) and that of white ball be denoted by P(W).

We have to find P(B).

Therefore, P(B) = n(total)/n(B)

                         = $\frac{9}{21}$

                  P(B) = $\frac{3}{7}$