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

Question

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

(A)  \frac{1}{4}

(B)  \frac{1}{3}

(C)  \frac{1}{2}

(D)  \frac{2}{3}

(E)  \frac{3}{4}

unverifiedvoracity · Accepted Answer

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.

CursoFDX · Answer

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) .

hsbinfy · Answer

very easy question. Straight D.

P(orange fish after addition )=1/3

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

Alex75PAris · Answer

It is not a 700 level question !

gracie · Answer

(o+k)/(s+2k)=1/2
s/o=2:1
p(s before addition)=2/3

msk0657 · Answer

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

Cedy92 · Answer

But very easy mate.... ^^

EgmatQuantExpert · Answer

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

Champagnereef · Answer

Thank you for your opinion, I think it is a very good idea

garima2709 · Answer

The ques asks " What was the probability of choosing a silver fish before any more fish were added?"

Paras96 · Answer

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

Let "o" and "s" be the no. of orange fish before adding 
As per the given scenario,
=>(o+k)/(o+k+s+2k)=1/3
=>3o+3k = s + 2k + o + k
=>2o = s

Required probability = s/(s+o) = 2o/(2o+o)=2/3

Hence D

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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