Assume that that fish tank contains 5 Fantails and \(n\) other fishes. The total fishes in that tank are \(n+5\)

A is the case that the first fish chosen will be a Fantail
B is the case that the second fish chosen will be a Fantail

(1) We have \(P(A) = \frac{5}{n+5}=\frac{1}{2} \implies n=5\). Since we know how many fishes are in that fish tank, we will know the probability to select 2 Fantails from that tank. Sufficient.

(2) Note that the probability that the second fish chosen is based on the first fish selected. Hence we have

\(P(B|A)=\frac{4}{n+4} = \frac{4}{9} \implies n=5\). Sufficient.

The answer is D
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A fish tank contains a number of fish, including 5 Fantails. If two fish are selected from the tank at random, what is the probability that both will be Fantails?

(1) The probability that the first fish chosen will be a Fantail is 1/2.

(2) The probability that the second fish chosen will be a Fantail is 4/9.

Solution:

Statement 1: This choice says that the no. of fish=5, The probability of Fantail=5/10=1/2. Sufficient.

Statement 2:
Let the no. of fish=x,
1.Prob of first (Fish)+ Prob of second (Fantail) Or
2. Prob of both(Fantail)
1.[(x/x+5)]*[(x-1)/(x+5-1)]=4/9.

Solving this, we get x^2-9x-16=0.
x is not an integer.(NOT POSSIBLE)

2. This gives x=5.(Only Possible Value).

Therefore the answer is Option D.

Correct me if I am wrong.

Let the other fishes be x.we hav 5+x fishes in total.

Probability of both fantails=5/(5+x)*4/(5+x)-1
=5/(5+x)*4/(4+x)-----(1)

Statement 1:
First chosen fish is fantail.
=>5/(5+x)=1/2
=>x=5
So total 15 fishes

From (1) Probability can be found out...Sufficient

Statement 2:

4/(4+x)=4/9
=> x=5 again

15 fishes in total
Probability can be found....Sufficient

Each alone sufficient

The question makes no mathematical sense. If you pick fish from a tank, and the probability you get a Fantail on the first pick is 1/2, then the probability you get a Fantail on the second pick is also 1/2, if you don't have any information at all about the first fish you picked. There is no logical reason the second fish should be more or less likely than the first to be a Fantail. The two statements are logically incompatible. That's the reason the correct solutions above are not getting an integer value using Statement 2 - the statement is wrong.

What the question designer means to say is: "If the first selected fish is a Fantail, then the probability the second selection is also a Fantail is 4/9". Then the question makes sense, and using either statement you have 10 fish in total, so you can answer the question.
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Edited as suggested. Thank you.
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I appreciate that this makes the question more useful to test takers reading the forum, but if you fix up mistakes in prep company questions, but still attribute the improved question to the prep company, that might mislead test takers who will purchase prep materials based on the questions they see posted here. Those test takers may think there are fewer errors in the original source than there really are. And the converse situation would also be problematic - if someone, say, posted one of my questions here, but changed it, I'd not want the question attributed to me, since I take a lot of care with the wording of my questions and wouldn't want that wording altered.

Probably the best policy in general is to post the original version of the question, along with the source, and then the corrected version immediately underneath, indicating that the question has been fixed, with an explanation why.
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Done.
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I would choose D.

First is fantail probability 1/2 implies it is 5/10
hence for second to be fantail: 5/10*4/9
Sufficient

Similarly ,
for statement 2 - it is sufficient as above.

Answer D

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