A pond contains 33 fish. Two are caught, tagged, and released back into the pond. After the tagged fish have had a chance to mingle with the others, eight fish are caught and released, one at a time. Assume that every fish in the pond is equally likely to be caught each time, regardless of which fish have been caught (and released) previously (this is not a realistic assumption for real fish in a real pond). The chance that among the fish caught in the second stage (after the tagging), at least four and at most eight were previously tagged is.

0.2424

Step-by-step explanation:

Total number of fish = 33

Total number of tagged fish = 2

Each fish has equal probability of being caught (remember that the focus is on the second stage of fish selection) and that probability is 1/33 = 0.0303

- When the question says at least 4 and at most 8 are tagged fish, it means that 4/8 or 1/2 or half of the fish selected were tagged fish. This gives us the figure to use in computation.

- So first of all, the probability that a fish caught is a tagged fish is (1/33) x 2

which is = 0.0303 x 2  =  0.0606

Now if 8 fish were caught and we want to know the probability that at least half of the eight fish (which is 4 fish) were tagged fish, what do we do?

- Multiply the probability of picking a tagged fish by 4; since each fish is replaced before the next selection.

0.0606 x 4  =  0.2424

f(x)=2/3x+4/3 is the answer because is whose power in the senate is equal to the power of the speaker of the house?

step-by-step explanation:

what statement? is there an image?