Question

A pond is freshly stocked with 6 brown trout and 18 lake trout. The first fisherman in the area catches and releases a trout. He catches another trout a while later. What is the probability that the fisherman caught lake trout each time?
A.1/16

B.1/8

C.5/92

D.5/96

Answer 1

Number of brown trout = 6

Number of lake trout = 18

Total number of trouts = $6+18=24$

Probability to catch a lake trout first time = $\frac{18}{24}$

As the fisherman let the trout go, so probability for the second time is =

$\frac{18}{24}$

Hence, the probability becomes=

$\frac{18}{24}\times\frac{18}{24}=\frac{324}{576}$

= $\frac{9}{16}$

The probability to find the brown trout is $\frac{1}{16}$

as , $\frac{6}{24}\times\frac{6}{24}=\frac{36}{576}$

= $\frac{6}{96}=\frac{1}{16}$

Answer 2

Answer:

9/16

Not given in the options.

Step-by-step explanation:

Probability is the number of possible outcome expressed as a fraction of the total outcome.

Number of brown trout = 6

Number of lake trout = 18

Total number of trout = 18 + 6 = 24

Probability of picking a brown trout = 6/24

Probability of picking a lake trout = 18/24

The probability of picking a lake trout twice (with replacement)

= 18/24 × 18/24

= 9/16

Not given in the options.