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) Number of red balls and green balls are equal.

Answer

N/a

Explanation

For `n=6` this experiment is a Bernaull's experiment. Here the success is the ball drawn from the urn is of mark `X`.
`p=10/25=2/5impliesq=-2/5=3/5`
`:.` The distribution `Z` is a binomial distribution with
`n=6,p=2/5` and `q=3/5`
`:.P(Z=r)=.^(n)C_(r)p^(r)q^(n-r)=.^(6)C_(r)(2/5)^(r)(3/5)^(6-r)`
(i) `P` (all balls are mark `X`)`=P` (6 successes)
`=.^(6)C_(6)p^(6)q^(0)=(2/5)^(6)`
(ii) `P` (more than 2 balls are mark Y)
`=P` (less than 4 balls are mark X)
`=P` (at least 4 successes)
`=P(4)+P(5)+P(6)`
`=.^(6)C_(4)p^(4)q^(2)+.^(6)C_(5)p^(5)q+.^(6)C_(6)p^(6)q^(0)`
`=.^(6)C_(4)(2/5)^(4)(3/5)^(2)+.^(6)C_(5)(2/5)^(5)(3/5)^(1)`
`+.^(6)C_(6)(2/5)^(6).(3/5)^(0)`
`=15xx(2/5)^(4)9/25+6xx(2/5)^(5)xx3/5+xx(2/5)^(6)xx1`
`=(2/5)^(4)[27/5+36/25+4/25]=(2/4)^(4)[(135+36+4)/25]`
`=(2/5)^(4)xx175/25=7(2/5)^(4)`
(iii) `P` (at least one ball is marked Y)
`=P` (at least 1 failure)
`=P` (at most 5 successes)
`=1-P(6)=1-.^(6)C_(6)p^(6)q^(0)`
`=1-1xx(2/5)^(6)xx1=1-(2/5)^(6)`
(iv) Required Probability
`=P`(3 successes and 3 failures)
`=P(3)=.^(6)C_(3)p^(3)q^(3)`
`=.^(6)C_(3)(2/5)^(3)(3/5)^(3)=20xx(2/5)^(3)xx(3/5)^(3)`
`=20xx8/15xx27/125=864/3125`
Required mena `=np=6xx2/5=2.4`