Probability doubt | Red, Green, Blue and Yellow Cards

[ronsom]

A deck of card contains 2 of each - Red, Green, Blue and Yellow cards.
If 2 cards are drawn randomly from the deck, what is the probability that they both are not blue?

I can think of 2 approaches to solve this problem, both of which give me a different result!

Approach (1):- 1 - P(both are blue)
P(both are blue) = 1st card is blue AND 2nd card is blue
= (2/8) * (1/7) = 1/28
1 - (1/28) = 27/28 <--- Incorrect answer


Approach (2):- P(both are either red or yellow or green)
P(R,G,Y) = (6/8) * (5/7)
= 15/28 <---- Correct answer



Can someone please point out what is it that I'm not doing correctly in Approach (1)?
Also, I believe both approaches should render the same answer... if not, which one to choose while answering a question?

[liferocks]

I think the question is 'neither is not blue'..this will include both are not blue and any one is not blue

or P(both are either red or yellow or green)
In the first approach it will be,

1 - [P(both are blue) +P(any one is blue)]
1-[1/28+2*{(2/8)*(6/7)}]
=15/28

[ronsom]

Thanks! that clears all doubts

[kstv]

A deck of card contains 2 of each - Red, Green, Blue and Yellow cards.
If 2 cards are drawn randomly from the deck, what is the probability that they both are not blue?
Approach (1):- 1 - P(both are blue)
P(both are blue) = 1st card is blue AND 2nd card is blue
= (2/8) * (1/7) = 1/28
1 - (1/28) = 27/28 <--- Incorrect answer

No of ways two cards can be drawn = 8C2 = 28
No of ways two cards that are not blue can be drawn = 6C2 = 15 <---- Correct method ?