Probabilités conditionnelles - Ex2
Question

Solution
Question 1
\(\mathrm P\left(A\cap B\right)=\mathrm P(A)\times\mathrm P_A(B)=0,4\times0,8\,\boxed{\color{red}=0,32}\)
Question 2
\(\mathrm P(B)=\dfrac{\mathrm P\left(A\cap B\right)}{\mathrm P_B(A)}=\dfrac{0,32}{0,6}=\dfrac{32}{60}\,\boxed{\color{red}=\dfrac{8}{15}}\)
Question 3
\(\mathrm P\left(A\cup B\right)=\mathrm P(A)+\mathrm P(B)-\mathrm P\left(A\cap B\right)=0,4+\frac{8}{15}-0,32\,\boxed{\color{red}=\frac{46}{75}}\)