A box contains 5 green, 8 yellow and 7 brown balls. One ball is taken out from a box at random. What is the probability that the ball taken out is (i) yellow ? (ii) brown? (ii) neither green nor brown ? (iv) not brown?
Number of green balls in the box = 5
Number of yellow balls in the box = 8
Number of brown balls in the box = 7
Total number of balls including green, yellow and brown in the box = 5 + 8 + 7 = 20
Now, let the event that a green ball is taken from the box be G.
the event that a yellow ball is taken from the box be Y.
the event that a green ball is taken from the box be B.
Number of favourable outcomes of event G = 5
Number of favourable outcomes of event Y = 8
Number of favourable outcomes of event B = 7
Total number of favourable outcomes = 5 + 8 + 7
= 20
(i) Probability of a yellow ball to be taken from the box = Probability of event Y occurring = P(Y)
P(Y) = 
(ii) Probability of a brown ball to be taken from the box = Probability of event B occurring = P(B)
P(B) = 
(iii) Probability of the taken ball is nether green nor brown = 1 – Probability of ball taken is either green or brown = 1- P(G or B)
P(G or B)’ = 1-P(G or B)
= 
= 
(iv) Probability of the ball taken is not brown = 1-Probability of the ball taken is brown = 1-P(B)
P(B’) = 1- P(B)
= 
= 
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