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You toss a fair coin 3x, events:

A = "first flip H"

B = "second flip T"

C = "all flips H"

D = "at least 2 flips T"

Q: Which events are independent?

From the informal def. it is where one doesnt affect the other.

So in this case, $AB$ seem independent? Any others?

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  • $\begingroup$ You are right about $A,B.$ It is impossible for both $B$ and $C$ to happen simultaneously, and those cannot be independent from each other. $A$ is a necessary condition for $C$ Regarding $D,$ if $A$ happens does that make $D$ more likely, less likely or unchanged. How about $B?$ And again $C$ and $D$ are mutually exclusive. $\endgroup$
    – Doug M
    Commented Feb 2, 2017 at 1:34

2 Answers 2

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If $X$ can be either head or tail, then:

$P(A)=P(HXX)=4/8\\ P(B)=P(XTX)=4/8\\ P(C)=P(HHH)=1/8\\ P(D)=P(HTT,THT,TTH,TTT)=4/8 $

So

$P(AB)=P(HTX)=2/8=P(A)P(B)$ (independent)

$P(AC)=P(HHH)=1/8\ne P(A)P(C)$ (not independent)

$P(AD)=P(HTT)=1/8\ne P(A)P(D)$ (not independent)

$P(BC)=0\ne P(B)P(C)$ (not independent)

$P(BD)=P(HTT,TTH,TTT)=3/8\ne P(B)P(D)$ (not independent)

$P(CD)=0\ne P(C)P(D)$ (not independent)

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The experiment can be represented by a uniformly distributed sample space of outcomes $$\rm\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

Of which we are considering the events:

$A = {\rm\{HHH, HHT, HTH, HTT\}} \\ B= {\rm\{HTH, HTT, TTH, TTT\}} \\\quad A\cap B={\{HTH,HTT\}} \\ C={\rm \{HHH\}} \\ \quad A\cap C=\{\} \\ \quad B\cap C=\{\} \\ D={\rm \{HTT, THT, TTH, TTT\}} \\ \quad A\cap D={\rm \{HTT\}} \\ \quad B\cap D={\rm\{HTT, TTH, TTT\}} \\ \qquad A\cap B\cap D ={\rm\{HTT\}} \\ \quad C\cap D = \{\}$

So only the events $A$ and $B$ are pairwise independent.

However, the events $A,B,D$ are triowise independent, as $\mathsf P(A\cap B\cap D)=\mathsf P(A)\mathsf P(B)\mathsf P(D)$, although they are not mutually independent.   (We could also say $A\cap B$ and $D$ are pairwise independent.)

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