Probabilités conditionnelles - Ex1
Question

Solution
Question 1

Question 2
\(\mathrm P_A(B)=\frac{\mathrm P\left(A\cap B\right)}{\mathrm P(A)}=\frac{0,3}{0,4}\,\boxed{\color{red}=\frac34}\)
\(\mathrm P_B\left(\overline{A}\right)=\frac{\mathrm P\left(\overline{A}\cap B\right)}{\mathrm P(B)}=\frac{0,1}{0,4}\,\boxed{\color{red}=\frac14}\)