If two marbles are picked at random, what is the probability that both marbles are red?

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

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

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

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