Bunuel wrote:

The hold of a fishing boat contains only cod, haddock, and halibut. If a fish is selected at random from the hold, what is the probability that it will be a halibut or a haddock?

(1) There are twice as many halibut as cod in the hold, and twice as many haddock as halibut.

(2) Cod account for 1/7 of the fish by number in the hold.

Asked to find the P (HA or HD)

Stat 1: If there is one cod(C) ..there are two Halibut(HB). C = 2HB.

then given twice as many haddcock(HD) as HB i.e. 4HB as it we know there are 2 HB...

C:HB:HD = 1:2:4 => Total 7 in number.

P (HB or HD) = 2/7 + 4/7 = 6/7...sufficient...

Stat 2: C = 1/7...we need to understand that probability of selecting of C is 1/7 then probability of selecting other two = 1 - 1/7 = 6/7..Sufficient..

IMO D is correct answer..

OA please...will correct if I missed anything..