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.

Answer 1

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.`