A committee of 3 members is to be selected to be selected out of 3 man and 2 women . What is the probability that the committee has at least one women?

n(s) = 50
Prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
? n(E) = 15
? p(E) = 15/50 = 3/10 = 0.3

n(S) = 49
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49

n(S) = 2³ = 8
Let E = Event of getting exactly two heads
= {(H, H, T), (H, T ,H), (T, H, H)}
? n(E) = 3
? required probability = 3/8

Total ways = 52
There is one queen of club and one king of heart.
? Favorable ways = 1 + 1 = 2
? Required probability = 2/52 = 1/26

Required probability = 3/52 + 13/52 = 16/52 =4/13
[Hint Why 13/52 because there are 13 spades and why 3/52 instead of 4/52 (there are four kings) because one king is already counted in spades.]

Number of ways to select 3 marbles out of 7 marbles = n(s) = ⁷C₃ = 35
Probability that 2 are green and 1 is red = n(E) = ⁴C₂ x ³C₁ = 18
? Required probability = 18/35

Total number of marbles = 6 + 4 + 2 + 3 = 15
Ways of selection of two red marbles = n(E) = ⁶C₂
Ways of selection of two marbles = n(S) = ¹⁵C₂
So, required probability = (⁶C₂) / (¹⁵C₂)
= (6 x 5) / (15 x 14) = 1/7

Ways of selection of two blue marbles = ⁴C₂
Ways of selection of one yellow marble = ³C₁
Ways of selection of three marbles = ¹⁵C₃
So, required probability = (⁴C₂ x ³C₁) / (¹⁵C₃)
= [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}]
= [(4 x 3 x 2 x 1) / (2 x 1 x 2 x 1)] x [(3 x 2 x 1) / (1 x 2 x 1)] / [ (15 x 14 x 13 x 12!) / (3 x 2 x 12!)]
= (6 x 3) / (5 x 7 x 13)
= 18 / 455

Probability that none is blue from four marbles
= ^{15 - 4}C₄ / ¹⁵C₄
= ¹¹C₄ / ¹⁵C₄ = 22/91

So, probability that at least one is blue from four marbles = 1 - 22/91 = 69/91

Ways of selection of two green marbles = ²C₂ = 1
Ways of selection of two yellow marbles = ³C₂ = 3
So, probability (both are green) = 1/¹⁵C₂ ....(I)
Probability (both are yellow) = 3/¹⁵C₂ ....(Ii)

Then, required probability = 1/¹⁵C₂ + 3/¹⁵C₂ = 4/ ¹⁵C₂
= 4/105