If four marbles are picked at random, what is the probability that at least one is blue?

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

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

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

Required probability = (²C₁ x ³C₂ + ²C₂ x ³C₁) / (⁵C₃) = 9/10

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

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

Ways of selection of 4 marbles = n(S) = ¹⁵C_{4
Ways of selection of one green marbles = n(E1) = ²C1
Ways of selection of two blue marbles = n(E2)=⁴C2
Ways of selection of one red marble = n(E3) = ⁶C1
? Required probability = (²C1 x ⁴C2 x ⁶C1) / (¹⁵C4) = 24/455}

Total number of caps = 2 + 4 + 5 + 1 = 12
Total number of outcomes = n(S) = ¹² C₂ = 66
Favorable number of outcomes = n(E) = ²C₂ = 1
? Required probability = 1/66

Total number of caps = 12
Total number of result = n(S) = ¹²C₄ = 495
Out of 5 caps, number of ways to not pick a green cap = n(E₁) = ⁵C₀ = 1
and out of 7 caps, number of ways to pic 4 caps = n(E) = ⁷C₄ = 35
? Required probability = (1 x 35)/495 = 7/99

Total number of caps = 12
? n(S) = ¹²C₃ = 220
n(E₁) = Out of 4 red caps, number of ways to pick 2 caps = ⁴C₂ = 6
n(E₂) = Out of 5 green caps
Number of ways to pick one cap = ⁵C₁ = 5
P(E) = n(E₁) x n(E₂)/n(S) = (6 x 5)/220 = 3/22