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A fish tank contains orange fish and silver fish. After k more orange

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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 10 Jun 2016, 05:07
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  55% (hard)

Question Stats:

64% (01:31) correct 36% (01:39) wrong based on 137 sessions

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A fish tank contains orange fish and silver fish. After k more orange fish and 2k more silver fish are added, the probability of choosing an orange fish at random is \(\frac{1}{3}\) . What was the probability of choosing a silver fish before any more fish were added?

(A) \(\frac{1}{4}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{1}{2}\)

(D) \(\frac{2}{3}\)

(E) \(\frac{3}{4}\)

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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 10 Jun 2016, 06:30
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AbdurRakib wrote:
A fish tank contains orange fish and silver fish. After k more orange fish and 2k more silver fish are added, the probability of choosing an orange fish at random is \(\frac{1}{3}\) . What was the probability of choosing a silver fish before any more fish were added?

(A) \(\frac{1}{4}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{1}{2}\)

(D) \(\frac{2}{3}\)

(E) \(\frac{3}{4}\)


Equation (1): \(o + k = 1x\)

Equation (2): \(s + 2k = 2x\)

Multiplying the first Equation by 2: \(2o + 2k = 2x\). Let´s call this last equation, Equation (3): \(2o + 2k = 2x\)

We can subtract Eq (3) \(-\) Eq (2): \(2o - s = 0\), and then \(s = 2o\).

Then, \(\frac{s}{s+o} = \frac{2o}{3o} = \frac{2}{3}\).

The correct answer is letter (D).
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 10 Jun 2016, 09:02
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very easy question. Straight D.

P(orange fish after addition )=1/3

P(silver fish after addition)=1-1/3 = 2/3
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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post Updated on: 10 Jun 2016, 11:15
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new additions of Orange Fish (OF) to Silver Fish (SF) to the fish tank are in the ratio \(\frac{k}{2k}\) = \(\frac{1}{2}\).

As the probability of choosing an OF is 1/3, there must 1 OF in 3 fishes or there must 1 OF for every 2 SF or the ratio of OF to SF = 1:2. So, even after adding fishes in the ratio of 1 OF for 2 SF, the ratio of OF:SF is still 1:2. This is only possible, if the original ratio of OF to SF is also 1:2.

So, if the probability of picking OF is 1/3, the probability of picking SF is 2/3 (both before and after addition of k OF and 2k SF).

So, D.

Hope that helps.

Originally posted by unverifiedvoracity on 10 Jun 2016, 09:26.
Last edited by unverifiedvoracity on 10 Jun 2016, 11:15, edited 1 time in total.
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 10 Jun 2016, 09:47
It is not a 700 level question !
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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 10 Jun 2016, 19:56
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(o+k)/(s+2k)=1/2
s/o=2:1
p(s before addition)=2/3
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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 15 Jun 2016, 13:28
hsbinfy wrote:
very easy question. Straight D.

P(orange fish after addition )=1/3

P(silver fish after addition)=1-1/3 = 2/3




The question is asking about the probability of silver fish BEFORE they were added...your approach is different..
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 24 Jun 2016, 08:45
msk0657 wrote:
hsbinfy wrote:
very easy question. Straight D.

P(orange fish after addition )=1/3

P(silver fish after addition)=1-1/3 = 2/3




The question is asking about the probability of silver fish BEFORE they were added...your approach is different..



But very easy mate.... ^^
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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New post 18 Jul 2018, 04:52

Solution



Given:

    • A fish tank contains orange fish and silver fish.
    • After k more orange fish and 2k more silver fish are added into the tank, the probability of choosing an orange fish at random is \(\frac{1}{3\).
}

To find:

    • The probability of choosing a silver fish before any more fish were added?

Approach and Working:

Let there were ‘x’ orange fish and ‘y’ silver fish in tank before some more fishes were added in the tank.

    • Thus, after adding k more orange fish and 2k more silver fish into the tank,
      o Total orange fish= x+k
      o Total silver fish= y+2k
      o Total fish in the tank= x+y+3k

    • Hence, the probability of choosing an orange fish= \(\frac{x+k}{(x+y+3k)}\)
      o \(\frac{x+k}{(x+y+3k)} = \frac{1}{3}\)
      o 3x+3k= x+y+3k
      o 2x= y

    • Thus, the probability of choosing a silver fish before any more fish were added = \(\frac{y}{x+y}\)
      o \(\frac{2x}{x+2x}= \frac{2x}{3x}= \frac{2}{3}\)


Hence, the correct answer is option D.

Answer: D
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Re: A fish tank contains orange fish and silver fish. After k more orange &nbs [#permalink] 18 Jul 2018, 04:52
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