The chief

The chief fisherman, Peter, estimates that if he uses four lines, then the probability of making a catch on one line is 0.7. If he uses five lines, then the probability of making a catch on any line is 0.6. If he uses six lines, the probability of making a catch on any line is 0.5. If the maximum objective is to catch at least four fish, how many lines should he use to maximize the probability of catching his goal?

Correct answer:

n =  6

Step-by-step explanation:

$$\begin{gathered} 4\dots 0.7 \\ 5\dots 0.6 \\ 6\dots 0.5 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot 0.7^4\cdot (1-0.7{)}^{4-4}=1\cdot 0.7^4\cdot 0.3^0=0.2401 \\ {p}_5=(\genfrac{}{}{0ex}{}{5}{4})\cdot 0.6^4\cdot (1-0.6{)}^{5-4}=5\cdot 0.6^4\cdot 0.4^1+\left(\genfrac{}{}{0ex}{}{5}{5}\right)\cdot 0.6^5\cdot (1-0.6{)}^{5-5}=1\cdot 0.6^5\cdot 0.4^0\doteq 0.337 \\ {p}_6=(\genfrac{}{}{0ex}{}{6}{4})\cdot 0.5^4\cdot (1-0.5{)}^{6-4}=15\cdot 0.5^4\cdot 0.5^2+\left(\genfrac{}{}{0ex}{}{6}{5}\right)\cdot 0.5^5\cdot (1-0.5{)}^{6-5}=6\cdot 0.5^5\cdot 0.5^1+(\genfrac{}{}{0ex}{}{6}{6})\cdot 0.5^6\cdot (1-0.5{)}^{6-6}=1\cdot 0.5^6\cdot 0.5^0\doteq 0.3438 \\ {p}_6>p_5>p_4 \\ n=6 \end{gathered}$$

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