A pond contains two types of fish: 10 blue and 9 red. Among red fish, $\frac{1}{4}$ are poisonous and among blue fish $\frac{3}{5}$ are poisonous. A woman and a man catch a fish. They cook and eat half the fish each. If you eat the poisonous fish, the odds of getting sick are $\frac{75}{100}$ for a man and $\frac{45}{100}$ for a woman. Calculate:
a) the probability that the woman will become ill(Pw)
b) the probability that at most one of the two gets sick
c) the probability that no one will get sick
d) the probability that they ate a poisonous blue fish if the man became ill
e) the probability that they ate a poisonous fish if no one got sick.
a)
Probability of catching a blue poisonous fish is Pb=$\frac{3}{5}\cdot 10$ the probability of catching a red poisonous fish is Pv$\frac{1}{4}\cdot 9$ and the probability of the woman getting sick is $\frac{45}{100}$ so we have Pw = $\frac{3}{5}\cdot 10 \cdot \frac{45}{100}$ + $\frac{1}{ 4}\cdot 10 \cdot \frac{45}{100}$
b)
Probability of a woman getting sick plus the probability of a man getting sick.
Pt = Pw + Pm = $\frac{3}{5}\cdot 10 \cdot \frac{45}{100}$ + $\frac{1}{ 4}\cdot 10 \cdot \frac{45}{100} + \frac{3}{5}\cdot 10 \cdot \frac{45}{100}$ + $\frac{1}{ 4}\cdot 10 \cdot \frac{75}{100}$
c) 1 - Pt
d)$\frac{\frac{75}{100} \cdot \frac{3}{4}\cdot10}{\frac{75}{100}}$
e) The probability that none got sick is $Pn=\frac{25}{100} \cdot \frac{55}{100}$.
The probability that they ate some poisonous fish is $Pp= \frac{1}{4} \cdot 9 + \frac{3}{5} \cdot 10$. So $P = \frac{Pn \cdot Pp}{Pn}$
Are these alternatives correct?
Thanks for the attention.