Question

An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

(i) all will bear ‘X’ mark.

(ii) not more than 2 will bear ‘Y’ mark.

(iii) at least one ball will bear ‘Y’ mark

(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Answer 1

Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

p = P (ball bearing mark ‘X’) =`10/25 = 2/5`

q = P (ball bearing mark ‘Y’) =`15/25 = 3/5`

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p = 2/5

∴ P (Z = z) = `""^nC_zp^(n-z)q^z`

(i) P (all will bear ‘X’ mark) = P (Z = 0) =`""^6C_0 (2/5)^6 = (2/5)^6`

(ii) P (not more than 2 bear ‘Y’ mark) = P (Z ≤ 2)

= P (Z = 0) + P (Z = 1) + P (Z = 2)

(iii) P (at least one ball bears ‘Y’ mark) = P (Z ≥ 1) = 1 − P (Z = 0)

`= 1 - (2/5)^6`

(iv) P (equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z = 3)

= C36 253 353

`= 20x8x27/15628`

`= 864/3125`