Here's the question you clicked on:
JonnyBaybee
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?
i got same, answer was 5 and it said it was incorrect.
i got answer 3 with Pr=.4091
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
Oh. Sorry. I didn't know that.
mubzz I sooo appreciate your help. I like Baye's Probability better haha it's easier to figure out
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!