Princeton Review
A fish tank contains a certain number of fishes, including 5 Fantails. If two fishes are selected from the tank at random, what is the probability that both will be Fantails?
1) The probability that the first fish have chosen will be a Fantail is 1/2.
2) If the first selected fish is a Fantail, then the probability the second selection is also a Fantail is 4/9.
OA D
A fish tank contains a certain number of fishes, including 5
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Let the other fishes be \(x\). We have \(5+x\) fishes in total.
Probability of both fantails \(=\frac{5}{5+x}*\frac{4}{5+x}-1\)
\(=\frac{5}{5+x}*\frac{4}{4+x}\cdots(1)\)
Statement 1:
First chosen fish is fantail.
\(\Rightarrow \frac{5}{5+x}=\frac{1}{2}\)
\(\Rightarrow x=5\).
So total \(15\) fishes.
From \((1)\) Probability can be found out... Sufficient
Statement 2:
\(\frac{4}{4+x}=\frac{4}{9}\)
\(\Rightarrow x=5\) again
\(15\) fishes in total
Probability can be found out.... Sufficient
Therefore, __D__ is the correct option.
Probability of both fantails \(=\frac{5}{5+x}*\frac{4}{5+x}-1\)
\(=\frac{5}{5+x}*\frac{4}{4+x}\cdots(1)\)
Statement 1:
First chosen fish is fantail.
\(\Rightarrow \frac{5}{5+x}=\frac{1}{2}\)
\(\Rightarrow x=5\).
So total \(15\) fishes.
From \((1)\) Probability can be found out... Sufficient
Statement 2:
\(\frac{4}{4+x}=\frac{4}{9}\)
\(\Rightarrow x=5\) again
\(15\) fishes in total
Probability can be found out.... Sufficient
Therefore, __D__ is the correct option.