rochioblogmz

2022-10-29

Complete the following probability table:

honotMornne

Step 1
Given data
$P\left(B\right)=0.70\phantom{\rule{0ex}{0ex}}P\left({B}^{C}\right)=0.30\phantom{\rule{0ex}{0ex}}\text{Total}=1\phantom{\rule{0ex}{0ex}}P\left(A|B\right)=0.10\phantom{\rule{0ex}{0ex}}P\left(A|{B}^{C}\right)=0.65\phantom{\rule{0ex}{0ex}}P\left(A\cap B\right)=P\left(B\right)×P\left(A|B\right)\phantom{\rule{0ex}{0ex}}=0.70×0.10=0.07\phantom{\rule{0ex}{0ex}}P\left(A\cap {B}^{C}\right)=P\left({B}^{C}\right)×P\left(A|{B}^{C}\right)\phantom{\rule{0ex}{0ex}}=0.30×0.65=0.195\phantom{\rule{0ex}{0ex}}P\left(A\right)=P\left(A\cap B\right)+P\left(A\cap {B}^{C}\right)\phantom{\rule{0ex}{0ex}}P\left(A\right)=0.07+0.195=0.265\phantom{\rule{0ex}{0ex}}P\left(B|A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{0.07}{0.265}=0.2641\phantom{\rule{0ex}{0ex}}P\left({B}^{C}|A\right)=\frac{P\left(A\cap {B}^{C}\right)}{P\left(A\right)}=\frac{0.195}{0.265}=0.7358\phantom{\rule{0ex}{0ex}}\text{Total}=P\left(B|AP\left({B}^{C}|A\right)=0.9999$

Do you have a similar question?