Nr. 5

$P(A)2,6\mathrm{\%}$
$P(A\cap B)=P(A)\ast P(B)=0,026\ast 0,1226\approx 0,0032=0,032\mathrm{\%}$
$P_A(B)=9,6\mathrm{\%}$
$P(\bar{A}\cap B)=P(\bar{A})\ast P(B)=1-P(A)\ast P(B)=0,974\ast 0,1226\approx 0,119=11,9\mathrm{\%}$
$P_B(A)=\frac{P(A\cap B)}{P(B)}=\frac{0,0032}{0,1226}=0,00036032=0,036\mathrm{\%}$

Nr. 14

a)

$P_E(F)=\frac{P(F\cap E)}{P(E)}=\frac{0,095}{0,113}=0,84=84\mathrm{\%}$

b)

$P_F(\bar{E})=\frac{P(F\cap \bar{E})}{P(F)}=\frac{0,005}{0,1}=0,05=5\mathrm{\%}$

c)

$P(E\cap \bar{F}{)}^2=0,{018}^2=0,000324=0,032\mathrm{\%}$
$P_{E^2}(\bar{F})=\frac{P(\bar{F}\cap E^2)}{P(E^2)}=$

Nr. 21

a)