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Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that the second toss results in heads, and let C be the event that in both tosses the coin lands on the same side. Show that the events A, B, and C are pairwise independent—that is, A and B are independent, A and C are independent, and B and C are independent—but not independent.

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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}

[tex]A\cap B=[/tex]{HH}

[tex]B\cap C[/tex]={HH}

[tex]A\cap C[/tex]={HH}

P(E)=[tex]\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}[/tex]

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

[tex]P(A)=\frac{2}{4}=\frac{1}{2}[/tex]

Number of favorable cases to event B=2

Number of favorable cases to event C=2

[tex]P(B)=\frac{2}{4}=\frac{1}{2}[/tex]

[tex]P(C)=\frac{2}{4}=\frac{1}{2}[/tex]

If the two events A and B are independent then

[tex]P(A)\cdot P(B)=P(A\cap B)[/tex]

[tex]P(A\cap)=\frac{1}{4}[/tex]

[tex]P(B\cap C)=\frac{1}{4}[/tex]

[tex]P(A\cap C)=\frac{1}{4}[/tex]

[tex]P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}[/tex]

[tex]P(B)\cdot P(C)=\frac{1}{4}[/tex]

[tex]P(A)\cdot P(C)=\frac{1}{4}[/tex]

[tex]P(A)\cdot P(B)=P(A\cap B)[/tex]

Therefore, A and B are independent

[tex]P(B)\cdot P(C)=P(B\cap C)[/tex]

Therefore, B and C are independent

[tex]P(A\cap C)=P(A)\cdot P(C)[/tex]

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

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