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- anonymous

Assume that
the fisherman can use between 3 and 6 lines. With 3 lines, the probability
of a catch on each line is 0.77. With 4 this probability is 0.62, with
5 it is 0.51, and with 6 the probability is 0.42.
How many lines should the fisherman use to maximize the probability of catching
at least 2 fish?

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- anonymous

- schrodinger

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- anonymous

its not 4 or 5

- anonymous

i got same, answer was 5 and it said it was incorrect.

- anonymous

i got answer 3 with Pr=.4091

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- anonymous

is that right?

- anonymous

Remember, in probability, each line will be treated as an individual event. The values calculated by @Calcmathlete are incorrect as the probability exceeds 1. Remember, for any given event the probability can NEVER be greater than 1 or less than 0.
In all these scenarios, he has to catch AT LEAST 2 fish so there can be 2 or more fish in each case.
So lets look at it case by case:
with 3 lines:
P (at least 2 fish) = P(3 fishes) or P(2 fishes)
P (at least 2 fish) = (0.77^3) + [(0.77^2) * 0.23]
P (at least 2 fish) = 0.5929
with 4 lines:
P (at least 2 fish) = P(4 fishes) or P(3 fishes) or P(2 fishes)
P (at least 2 fish) = (0.62^4) + [(0.62^3) * 0.38] + [0.62^2 + 0.38^2]
P (at least 2 fish) = 0.2938
with 4 lines:
P (at least 2 fish) = P(5 fishes) or P(4 fishes) or P(3 fishes) or P(2 fishes)
P (at least 2 fish) = (0.51^5) + [(0.51^4) * 0.49)] + [(0.51^3) * (0.49^2)] + [0.51^2 + 0.49^3]
P (at least 2 fish) = 0.1301
It is obvious from here that with 6 lines, this probability will decrease. So you use 3 lines. The probability will be 0.5929

- anonymous

Oh. Sorry. I didn't know that.

- anonymous

mubzz I sooo appreciate your help. I like Baye's Probability better haha it's easier to figure out

- anonymous

Haha yea there are definitely several ways to approach any given problem. As long as you stick to the principles, no restriction on what method you use :D Good luck!

- anonymous

Thank You!

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