Answer with Explanation:
We are given that
E1,E2,E3 and E4 are equally likely outcomes.
Therefore, the probability
[tex]P(E_1)=P(E_2)=P(E_3)=P(E_4)[/tex]
We know that
Sum of all probabilities=1
[tex]P(E_1)+P(E_2)+P(E_3)+P(E_4)=1[/tex]
[tex]P(E_1)+P(E_1+P(E_1)+P(E_1)=1[/tex]
[tex]4P(E_1)=1[/tex]
[tex]P(E_1)=\frac{1}{4}[/tex]
a.[tex]P(E_2)[/tex]=[tex]\frac{1}{4}[/tex]
b.Favorable cases=2
Total outcomes=4
Probability,[tex]P(E)=\frac{favorable\;cases}{Total\;number\;of\;cases}[/tex]
The probability that any two of the outcomes occur=[tex]\frac{2}{4}=\frac{1}{2}[/tex]
c.Favorable cases=3
[tex]P(E)=\frac{3}{4}[/tex]
Answer:
outcomes common to E1 and E2 = [1, 3, 5, 7, 9, 19, 21, 23, 25, 27]
Explanation:
PLATO
