If three caps are picked at random, what is the probability that two are red and one is green?

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 = 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

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}

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

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

Required probability = 1 - ⁸C₂/ ¹²C₂
= 1 - 28/66 = 38/66 = 19/33

Total number of possible outcomes
= n(S) = ¹²C₂ = (12 x 11)/2 = 66

Total number of favorable events
= n(E) = ⁴C₂
= (4 x 3) / (2 x 1) = 6
? Required probability = 6/66 = 1/11

Total Number of outcomes = ¹⁰C₃ = 120
Number of outcomes not containing red balls = ⁵C₃ = 10
? Probability that at least one is red = 1 - 10/120 = 11/12

Total number of outcomes = ¹⁰c₄ = 210
Favourable number of outcomes = ³c₂ x ²c₂ = 3 x 1 = 3
? Required probability = 3/210 = 1/70

n(S) = Number of ways of drawing 2 cards out of 52
⁵²C₂ = (52 x 51) / (2 x 1) = 1326
n(E) = Number of ways of drawing 2 cards out of 4
= ⁴C₂ = (4 x 3)/2 = 6
? P(E) = 6/1326 = 1/221