M1 probability

1.
Let 9n be the number of the group of students.

Number of students from class A = (5/9)9n = 5n
Hence, P(class A) = (5n/9n) = = (5/9)

Number of boys in class A = 5n x 60% = 3n
Number of girls in class A = 5n - 3n = 2n

Number of students in class A that perform drama at Christmas
= 3n x 30% + 2n x 75%
= 0.9n + 1.5n
= 2.4n
Hence, P([class A] and [perform drama]) = 2.4n/9 = 2.4/9

The required probability
= P([perform drama] | [class A])
= P([cass A] and [perform drama]) / P(class A)
= (2.4n/9) / (5n/9)
= 0.48


2.
B : black ball
N : not black ball

P(B) = 5 / (5 + 2 + 4) = 5/11
P(N) = 1 - (5/11) = 6/11

P(B) = 5/11
P(2nd ball is B) = P(NB) = (6/11)x(5/11)
P(3rd ball is B) = P(NNB) = (6/11)²x(5/11)
........
P(nth ball is B) = (6/11)ⁿ⁻¹x(5/11)

P(at least one black ball in n draws) > 0.9
(5/11) + (6/11)x(5/11) + (6/11)²x(5/11) + ...... + (6/11)ⁿ⁻¹x(5/11) > 0.9
(5/11) x [1 + (6/11) + (6/11)² + ..... + (6/11)ⁿ⁻¹] > 0.9
1 + (6/11) + (6/11)² + ..... + (6/11)ⁿ⁻¹ > 0.9 x (11/5)
1 + (6/11) + (6/11)² + ..... + (6/11)ⁿ⁻¹ > 1.98

Sum of G.P. = a(1 - rⁿ)/(1 - r)
Hence, 1 x [1 - (6/11)ⁿ] / [1 - (6/11)] > 1.98
1 - (6/11)ⁿ > 1.98 x (5/11)
1 - (6/11)ⁿ > 0.9
(6/11)ⁿ < 0.1

Since log(0.1) < 0
then, log(6/11)ⁿ > log(0.1)
n[log(6) - log(11)] > log(0.1)
n > log(0.1) / [log(6) - log(11)]
n > 3.799

Hence, the least value of n = 4