Problem Solving

Q). In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
A) 8/25
B) 12/35
C) 13/35
D) 9/25
E) 17/25

OA is B

I did the question in this way and was not able to get to the right answer:-
I found the probability of getting a nickel in the first 2 shots and subtract it from 1
i.e.
1st pick: 6/15
2nd pick: 5/14

therefore, 6*5/(15*14)
-> 1/7
therefore probability of not getting a nickel is 1- 1/7=> 6/7
But I see that I'm doing something wrong here.

So,I had to move back to the conventional method in order to get the answer
i.e.
P(D)*P(D) + P(P)*P(P) +2 P(D)P(P)

Can someone explain me, where I went wrong with my earlier approach.

~Thanks,
Akshat

We have 4 possible outcomes if we draw two coins:

Nickel-nickel
Nickel-something else
Something else-nickel
Something else-something else

(Since the question asks specifically about nickels, it does not matter to us if that something else is a penny or a dime. Giving us more variables than we actually need is a common probability trap; always try and simplify if possible.)

The question asks for the probability of getting something other than a nickel on both draws, so we're looking at the 4th outcome:

P(SE-SE) = 9/15 * 8/14 = 3/5 * 4/7 = 12/35