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Question
An urn contains 25 balls of which 10 balls are red and the remaining green. A ball is drawn at random from the urn, the colour is noted and the ball is replaced. If 6 balls are drawn in this way, find the probability that:
(i) All the balls are red.
(ii) Not more than 2 balls are green.
(iii) The number of red balls and green balls is equal.
Solution
Number of red balls = 10
Number of Green balls = 25 - 10 = 15
Total number of balls = 25
Number of balls drawn is 6 with replacement
(i) P(all the balls are red) = `10/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25 = 64/15625` [Fixed case]
(ii) P(not more than 2 balls are green)
=P(2 green balls and 4 red balls) + P(1 green ball and 5 red balls) + P(all 6 red balls)
= P(first two green balls and next four red balls) × `(6!)/(2!4!)`
+ P(first green balls and next 5 red balls) × `(6!)/(5!)` + P(all 6 red balls)
`= (15/25 xx 15/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25 xx (6!)/(2!4!)) + (15/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25 xx (6!)/(1!5!)) + (10/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25 xx 10/25) = 112/625`
(iii) P(number of red balls and green balls are equal)
= P(3 red balls and 3 green balls)
= P(First three red balls and next three green balls)`xx (6!)/(3!3!)`
`= 10/25 xx 10/25 xx 10/25 xx 15/25 xx 15/25 xx 15/25 xx (6!)/(3!3!) = 864/3125.`
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