Answer with Step-by-step explanation:
We are given that two independent tosses of a fair coin.
Sample space={HH,HT,TH,TT}
We have to find that A, B and C are pairwise independent.
According to question
A={HH,HT}
B={HH,TH}
C={TT,HH}
{HH}
={HH}
={HH}
P(E)=![\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%5C%3Bof%5C%3Bfavorable%5C%3Bcases%7D%7Btotal%5C%3Bnumber%5C%3Bof%5C%3Bcases%7D)
Using the formula
Then, we get
Total number of cases=4
Number of favorable cases to event A=2
![P(A)=\frac{2}{4}=\frac{1}{2}](https://tex.z-dn.net/?f=P%28A%29%3D%5Cfrac%7B2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Number of favorable cases to event B=2
Number of favorable cases to event C=2
![P(B)=\frac{2}{4}=\frac{1}{2}](https://tex.z-dn.net/?f=P%28B%29%3D%5Cfrac%7B2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B2%7D)
![P(C)=\frac{2}{4}=\frac{1}{2}](https://tex.z-dn.net/?f=P%28C%29%3D%5Cfrac%7B2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B2%7D)
If the two events A and B are independent then
![P(A)\cdot P(B)=P(A\cap B)](https://tex.z-dn.net/?f=P%28A%29%5Ccdot%20P%28B%29%3DP%28A%5Ccap%20B%29)
![P(A\cap)=\frac{1}{4}](https://tex.z-dn.net/?f=P%28A%5Ccap%29%3D%5Cfrac%7B1%7D%7B4%7D)
![P(B\cap C)=\frac{1}{4}](https://tex.z-dn.net/?f=P%28B%5Ccap%20C%29%3D%5Cfrac%7B1%7D%7B4%7D)
![P(A\cap C)=\frac{1}{4}](https://tex.z-dn.net/?f=P%28A%5Ccap%20C%29%3D%5Cfrac%7B1%7D%7B4%7D)
![P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}](https://tex.z-dn.net/?f=P%28A%29%5Ccdot%20P%28B%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B4%7D)
![P(B)\cdot P(C)=\frac{1}{4}](https://tex.z-dn.net/?f=P%28B%29%5Ccdot%20P%28C%29%3D%5Cfrac%7B1%7D%7B4%7D)
![P(A)\cdot P(C)=\frac{1}{4}](https://tex.z-dn.net/?f=P%28A%29%5Ccdot%20P%28C%29%3D%5Cfrac%7B1%7D%7B4%7D)
![P(A)\cdot P(B)=P(A\cap B)](https://tex.z-dn.net/?f=P%28A%29%5Ccdot%20P%28B%29%3DP%28A%5Ccap%20B%29)
Therefore, A and B are independent
![P(B)\cdot P(C)=P(B\cap C)](https://tex.z-dn.net/?f=P%28B%29%5Ccdot%20P%28C%29%3DP%28B%5Ccap%20C%29)
Therefore, B and C are independent
![P(A\cap C)=P(A)\cdot P(C)](https://tex.z-dn.net/?f=P%28A%5Ccap%20C%29%3DP%28A%29%5Ccdot%20P%28C%29)
Therefore, A and C are independent.
Hence, A, B and C are pairwise independent.