Question

A bag contains 4 white ball, 6 red balls, 7 black balls and 3 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) white
(ii) not black
(iii) neither white nor black

Answer 1

Answer:

$\huge\fcolorbox{black}{pink}{Answer-}$

A bag contains :

white balls = 4

red balls = 6

black balls = 7

blue balls = 3

________________

total balls = 20

1.) P (white ) = 4 / 20

= 1 / 5

2.) P ( not black) = 13 / 20

3.) P ( neither white nor black ) = 9 / 20

Answer 2

Given

$\bold{Total \ no: \ of \ white \ balls} = 4\\\bold{Total \ no: \ of \ red \ balls} = 6\\\bold{Total \ no: \ of \ black \ balls} = 7\\\bold{Total \ no: \ of \ blue \ balls} = 3\\\implies \bold{Total \ no: \ of \ balls} = 4 + 6 + 7 + 3 =\underline{\underline{20}}$

$(i) \bold{ Probability \ of \ getting \ a \ white \ ball} = \boxed{\boxed{\bold{\frac{4}{20}= \frac{1}{5} }}}}$

$(ii)\bold{ Probability \ of \ not \ getting \ black} = 1-\dfrac{7}{20} = \boxed{\boxed{\bold{\dfrac{13}{20}}}}$

$(iii) \bold{ Probability \ of \ getting \ neither \ white \ or \ black} = \boxed{\boxed{\bold{\dfrac{9}{20}}}}$