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The hold of a fishing boat contains only cod, haddock, and halibut. 28 Jul 2016, 05:23
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D
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70% (01:25) correct 30% (01:33) wrong based on 344 sessions

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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.
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D
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The hold of a fishing boat contains only cod, haddock, and halibut. 28 Jul 2016, 06:06
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get D

1)ratio of all 3 fish given
so probablity=6/7 for Halibut and haddock

2)same here
1-1/7=6/7 for th answer
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The hold of a fishing boat contains only cod, haddock, and halibut. 28 Jul 2016, 07:01
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Bunuel
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.
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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..
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The hold of a fishing boat contains only cod, haddock, and halibut. 28 Jul 2016, 07:29
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Bunuel
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?
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(1) There are twice as many halibut as cod in the hold, and twice as many haddock as halibut.

\(2c = l\\\\
2l = a\\ \\
\text{Total} = c + l + a = 7c\)

\(c = \frac{1}{7}\)

Sufficient (see below)

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

\(\Pr{(\text{halibut }\cap \text{ haddock})} = 1 - \Pr{(\text{cod})}\)

\(\Pr{(\text{halibut }\cap \text{ haddock})} = \frac{6}{7}\)

Sufficient

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(D) each statement alone is sufficient
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The hold of a fishing boat contains only cod, haddock, and halibut. 04 Aug 2016, 04:35
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Bunuel
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.
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(1) There are twice as many halibut as cod in the hold, and twice as many haddock as halibut.
Cod = x
Halibat =2x
Haddock = 4x

Probablity of haddock or halibut = 6x/7x = 6/7
SUFFICIENT

(2) Cod account for 1/7 of the fish by number in the hold.
Cod = 1/7
Halibut and haddock = 6/7
total = 1/7+6/7= 7/7 = 1
Probablity of picking haddock and halibut = 6/7/1 = 6/7

SUFFICIENT
ANSWER IS D
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The hold of a fishing boat contains only cod, haddock, and halibut. 11 Feb 2021, 18:24
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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.
Halibut = 2xCod
Haddock = 4xCod

4C + 2C + C

P(Haddock or halibut) = 6/7

Sufficient.

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

P(haddock or halibut) = 1 - 1/7 = 6/7

Sufficient.

Answer is C.
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The hold of a fishing boat contains only cod, haddock, and halibut. 25 Feb 2021, 03:53
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Answer is D Not C
CEdward
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.
Halibut = 2xCod
Haddock = 4xCod

4C + 2C + C

P(Haddock or halibut) = 6/7

Sufficient.

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

P(haddock or halibut) = 1 - 1/7 = 6/7

Sufficient.

Answer is C.
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The hold of a fishing boat contains only cod, haddock, and halibut. 11 Apr 2023, 09:01
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