Answer:
a.Yes, Event E and F are independent.
b.[tex]\frac{1}{4}[/tex]
Step-by-step explanation:
We are given that a coin is tossed twice.
S={HH,HT,TH,TT}
E={HH,HT}
F={HH,TH}
a.We have to find that event A and event B are independent or not.
We know that when two events A and B are independent then
[tex]P(A)\cdot P(B)=P(A\cap B)[/tex]
Probability,P(E)=[tex]\frac{number\;of\;favorable\;cases}{total\;number\;of cases}[/tex]
Total number of cases=4
P(E)=[tex]\frac{2}{4}=\frac{1}{2}[/tex]
P(F)=[tex]\frac{2}{4}=\frac{1}{2}[/tex]
[tex]E\cap F[/tex]={HH}
[tex]P(E\cap F)=\frac{1}{4}[/tex]
[tex]P(E)\cdot P(F)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}[/tex]
[tex]\P(E)\cdot P(F)=P(E\cap F)[/tex]
Therefore, Event E and event B are independent.
b.We have to find the probability of showing heads on both toss.
Number of favorable cases={HH}=1
Total number of cases=4
By using the formula of probability
The probability of getting heads on both toss=[tex]\frac{1}{4}[/tex]
