Question

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

Answer 1

Answer :-

The probability that

i.) the ball drawn is black is $\dfrac{3}{7}$

ii.) the ball drawn is white is $\dfrac{4}{7}$

Given :-

• Total number of black balls = 9

• Total number of white balls = 12

Formula to be used :-

$\boxed{\tt{Probability\: of \:\:an \:\:event = \dfrac{Number\:\: of\:\: favourable\:\: outcomes}{Total\:\: number\:\: of \:\:possible\:\: outcomes}}}$

Solution :-

Total number of possible outcomes

= Sum of all balls

= Total number of black balls + Total number of white balls

= 9 + 12

= 21

i.) If a black ball is drawn

Number of favourable outcomes = Number of black balls = 9

$\tt{Probability\: (the\:\: ball\:\: drawn\:\: is \:\:black)= \dfrac{Number\:\: of\:\: black\:\: balls}{Total\:\: number\:\: of \:\:possible\:\: outcomes}= \dfrac{9}{21}= \dfrac{3}{7}}$

ii.) If a white ball is drawn

Number of favourable outcomes = Number of white balls = 12

$\tt{Probability\: (the\:\: ball\:\: drawn\:\: is \:\:white)= \dfrac{Number\:\: of\:\: white\:\: balls}{Total\:\: number\:\: of \:\:possible\:\: outcomes}= \dfrac{12}{21}= \dfrac{4}{7}}$