pkw209 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 1/3. What was the probability of choosing a silver fish before any more fish were added?

a) 1/4

b) 1/3

c) 1/2

d) 2/3

e) 3/4

IMO D

From the info given, we infer

o+k/(o+s+3k)= 1/3
Cross multiplying and solving
s=2o

Now calculate prob of s/(o+s)

2s/(s/2 + s)

= 2/3

Hence D

Agree with eaakbari ans should be [spoiler]D)2/3[/spoiler]
just want to share my approach

since probability of choosing orange fish is 1/3 let us consider number of orange fish is 100 and total fish is 300

before adding fish ,number of orange fish is 100-k and total number is 300-k

so probability of selecting orange fish is (100-k)/(300-3k) or 1/3
hence probability of selecting silver fish {1-(1/3)} or 2/3

The probabilty of choosing orange or silver fish is 1.
If probability of one is given here it is 1/3, the other must be 1 - 1/3 = 2/3.

hachette wrote:The probabilty of choosing orange or silver fish is 1.
If probability of one is given here it is 1/3, the other must be 1 - 1/3 = 2/3.

can you please explain why in the approach you have taken 1/3 as the probability of choosing orange fish BEFORE 1/3 as in question it is mentioned as AFTER . is it something we can infer ?

pkw209 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 1/3. What was the probability of choosing a silver fish before any more fish were added

After the addition ,
assume no of orange fish = x silver fish is 2x
Prob(O)= x/x+2x = 1/3

Before addition,
Orange fish = x-k Silver fish = 2x-2k = 2(x-k)
Prob (S) = 2(x-k)/{(x-k) + 2 (x-k)} = 2(x-k)/3(x-k)