IB Maths AA HL Ahl 4 13 Hl Bayes Theorem Questions

METHOD 1 recognizing conditional probability
$\mathrm{P}\left(G^{\prime} \mid p o s\right)$ OR $\mathrm{P}(G \mid p o s)$$\frac{0.02(1-p)}{0.95 p+0.02(1-p)}\left(=\frac{18}{150}\right)$ OR $\frac{0.95 p}{0.95 p+0.02(1-p)}\left(=\frac{132}{150}\right)$

Note: Award A1 for a correct numerator and A1 for a correct denominator.
p=0.133738
p=0.134

METHOD 2
attempt to set up a system of equations ( S= sample size)
p(0.95 S)=132 and (1-p)(0.02 S)=18
attempt to solve for p or S
$\frac{0.95 p}{0.02(1-p)}=\frac{132}{18}$
OR $S=p S+(1-p) S=\frac{132}{0.95}+\frac{18}{0.02}=138.947 \ldots+900=1038.94 \ldots$$p=0.133738 \ldots$
p=0.134

METHOD 3
attempt to find the number of parrots with the gene and the number without number of parrots with the gene $\approx \frac{132}{0.95}=138.947 .$. AND
number of parrots without the gene $\approx \frac{18}{0.02}=900$
number of parrots in the sample $\approx 138.947 \ldots+900=1038.94 \ldots$
attempt to find proportion of sample with the gene
$p \approx \frac{138.947 \ldots}{1038.94 \ldots}=0.133738 \ldots$
p=0.134