Consider two independent tosses of a fair coin. Let Abe the event that the first toss results in heads, let Bbe the event that the second toss results in heads, and let Cbe 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.

Short Answer

Expert verified

All the all three events are not independent:

P(ABC)=P({(H,H)})=14121212=P(A)P(B)P(C)

Step by step solution

01

Step 1:Given Information

Given that two independent tosses of a fair coin. Let Abe the event that the first toss results in heads, let Bbe 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.

02

Step 2:Explanation

On the off chance that a fair coin is thrown twice freely there are 4 similarly reasonable events:

(H,H)- both tosses resulted in heads,P(H,H)=P(H)P(H)=1212=14

(H,T)- first toss is heads, the second tails,P(H,T)=14

(H,T)- first toss is tails, the second headsP(T,H)=14

(T,T)- both tosses resulted in tails,P(T,T)=14

These events are all mutually exclusive.

03

Explanation of Defined Events

Defined events:

P(A)=P({(H,H),(H,T)})=P({(H,H)})+P({(H,T)})=214=12

P(B)=P({(H,H),(T,H)})=12

P(C)=P({(H,H),(T,T)})=12

04

Step 4:Explanation of Characterization of Independence

Characterization of independence is that the probability of intersection is the result of the probabilities:

P(AB)=P({(H,H)})=14=1212=P(A)P(B)AandBare independent

P(BC)=P({(H,H)})=14=1212=P(B)P(C)BandCare independent

P(AC)=P({(H,H)})=14=1212=P(A)P(C)AandCare independent

All events are independent in pairs.

05

Step 5:Final Answer

Thus all three events are not independent:

P(ABC)=P({(H,H)})=14121212=P(A)P(B)P(C)

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