Answer:probability of an event E,
P(E)=No. of favourable outcomes[tex]\frac{x}{y}[/tex] total no. of outcomes
P(1)=4[tex]\frac{x}{y}[/tex]20=1[tex]\frac{x}{y}[/tex]5=0.20
P(2)=6[tex]\frac{x}{y}[/tex]20=3[tex]\frac{x}{y}[/tex]10=0.30
P(3)=2[tex]\frac{x}{y}[/tex]20=1[tex]\frac{x}{y}[/tex]10=0.10
P(4)=6[tex]\frac{x}{y}[/tex]20=3[tex]\frac{x}{y}[/tex]10=0.30
P(5)=2[tex]\frac{x}{y}[/tex]20=1[tex]\frac{x}{y}[/tex]10=0.10
Step-by-step explanation:
