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Question

A box contains 4 white balls, 6 red ball, 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) neither white nor black
(ii) red or white, and
(iii) either white or red or black or blue

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Solution

Given, white ball =4
Red ball =6
Black ball =7
Blue ball =3
Total number of balls =20
Therefore, n(S) =20
(i) Let A be event of getting neither White nor Black ball.
n(A)=6+3=9
P(A) =n(A)n(S)=920
(ii) Let B be event of getting Red or White ball.
n(B)=6+4=10
P(B)=n(B)n(S)=1020=12
(iii) Let C be event of getting either White or Red or black or blue ball.
n(C)=4+6+7+3=20
P(C)=n(c)n(s)=2020=1

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