Answer with explanation:
The four Options provided are
A. Getting exactly 1 head and Getting at least 2 tails
B. Getting at least 1 head and Getting at least 1 tail
C. Getting exactly 1 head and Getting at least 1 tail
D.Getting at least 2 heads and Getting at least 2 tails
→When Roger will flip a coin three times,
Total Possible Outcome
[tex]=2^3=8[/tex]
={H H H, H H T, H T H, T H H,H T T, T H T,T T H,T T T}
⇒Two events are said to be Disjoint, if
P (A ∩ B)=0→→P(A∪B)=P(A) +P(B)
Now, we will start from option ,1 that among four options which events are Disjoint.
1.A=Probability of Getting Exactly 1 Head={ H T T, T H T,T T H}
[tex]=\frac{3}{8}[/tex]
B=Probability of Getting at least 2 tails={H T T, T H T,T T H,T T T}
[tex]=\frac{4}{8}\\\\=\frac{1}{2}[/tex]
A ∩ B={H T T, T H T,T T H}≠0
→P (A∩B)≠0,
So, the two events are not Disjoint.
2.A=Probability of Getting At least 1 Head= ={H H H, H H T, H T H, T H H,H T T, T H T,T T H}
[tex]=\frac{7}{8}[/tex]
B=Probability of Getting At least 1 tail ={T T T, H H T, H T H, T H H,H T T, T H T,T T H}
[tex]=\frac{7}{8}[/tex]
A ∩ B={H H T, H T H, T H H,H T T, T H T,T T H}≠0
→P (A∩B)≠0,
So, the two events are not Disjoint.
3.A= Probability of Getting exactly 1 head={H T T, T H T,T T H}
B=Probability of Getting at least 1 tail={T T T, H H T, H T H, T H H,H T T, T H T,T T H}
A ∩ B={H T T, T H T,T T H}=3≠0
⇒P (A∩B)≠0,
So, the two events are not Disjoint.
4. A=Probability of Getting at least 2 heads ={H H H, H H T, H T H, T H H}
[tex]=\frac{4}{8}\\\\=\frac{1}{2}[/tex]
B=Probability of Getting at least 2 tails ={H T T, T H T,T T H,T T T}
[tex]=\frac{4}{8}\\\\=\frac{1}{2}[/tex]
A ∩ B=0
,⇒P (A∩B)=0,
So, the two events are Disjoint.
Option D: Getting at least 2 heads and Getting at least 2 tails, is disjoint event.
→There is no need to find Probability.You can write directly,that is if , A ∩ B≠0, the events are not disjoint.
