Question

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.

Answer 1

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}

`A\cap B=`{HH}

`B\cap C`={HH}

`A\cap C`={HH}

P(E)=
`(number\;of\;favorable\;cases)/(total\;number\;of\;cases)`

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

`P(A)=(2)/(4)=(1)/(2)`

Number of favorable cases to event B=2

Number of favorable cases to event C=2

`P(B)=(2)/(4)=(1)/(2)`

`P(C)=(2)/(4)=(1)/(2)`

If the two events A and B are independent then

`P(A)\cdot P(B)=P(A\cap B)`

`P(A\cap)=(1)/(4)`

`P(B\cap C)=(1)/(4)`

`P(A\cap C)=(1)/(4)`

`P(A)\cdot P(B)=(1)/(2)\cdot (1)/(2)=(1)/(4)`

`P(B)\cdot P(C)=(1)/(4)`

`P(A)\cdot P(C)=(1)/(4)`

`P(A)\cdot P(B)=P(A\cap B)`

Therefore, A and B are independent

`P(B)\cdot P(C)=P(B\cap C)`

Therefore, B and C are independent

`P(A\cap C)=P(A)\cdot P(C)`

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.