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# The hold of a fishing boat contains only cod, haddock, and halibut.

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The hold of a fishing boat contains only cod, haddock, and halibut. [#permalink]
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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.

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

$$c = \frac{1}{7}$$

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

(D) each statement alone is sufficient
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Re: The hold of a fishing boat contains only cod, haddock, and halibut. [#permalink]
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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.

(1) There are twice as many halibut as cod in the hold, and twice as many haddock as halibut.
Cod = x
Halibat =2x

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
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Re: The hold of a fishing boat contains only cod, haddock, and halibut. [#permalink]
1
Kudos
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

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.

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Re: The hold of a fishing boat contains only cod, haddock, and halibut. [#permalink]
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Answer is D Not C
CEdward 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.
Halibut = 2xCod

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.

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Re: The hold of a fishing boat contains only cod, haddock, and halibut. [#permalink]
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Re: The hold of a fishing boat contains only cod, haddock, and halibut. [#permalink]
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