In a deck of 52 cards there are 2 colours. Each colour having 26 cards. 

As we need to choose 4 cards out of 52. Let S represents the sample space. ∴ 

n(S) = ⁵²C₄ 

Let A represents the event that all 4 cards drawn are black. 

∴ n(A) = ways in which 4 cards can be selected from 26 black cards. 

⇒ n(A) = ²⁶C₄ 

∴ P(A) = \(\frac{^{26}C_4}{^{52}C_4}\)  = \(\frac{26\times25\times24\times23}{52\times51\times50\times49}\)  = \(\frac{46}{833}\)

Let B represents the event that all 4 cards drawn are red. 

∴ n(B) = ways in which 4 cards can be selected from 26 red cards. 

⇒ n(B) = ²⁶C₄ 

∴ P(B) =  \(\frac{^{26}C_4}{^{52}C_4}\)  = \(\frac{26\times25\times24\times23}{52\times51\times50\times49}\)  = \(\frac{46}{833}\)

As we need to find the probability of event such that all drawn cards are from same colour. This means we need to find 

P(A∪B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E∪F) = P(E) + P(F) – P(E∩F) 

∴ P(A∪B) = P(A) + P(B) – P(A∩B) 

As both events A and B have no common elements or we can say that they are mutually exclusive 

∴ P(A∩B) = 0 

Hence, 

P(A∪B) = P(A) + P(B) = \(\frac{46}{833} + \frac{46}{833} = \frac{92}{833}\)