the network of friendship, the question of probability theory (Grimmett and Stirzaker)

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Anne, Betty, Chloe, and Daisy were all friends at school. Subsequently each of the ${4 \choose 2} = 6$ subpairs meet up, at each of the six meetings the pair involved quarrel with some fixed probability $p$, or become firm friends with probability $1-p$. Quarrels take place independently of each other. In future, if any of the four hears a rumor, then she tells it to her firm friends only. If Anne hears a rumour, what is the probability that:

(d) Daisy hears it if she has quarreled with Anne?

Solution: The network of friendship is best represented as a square with diagonals with the concerns labelled $A$, $B$, $C$, and $D$. Draw a diagram. Each link of the network is absent with probability $p$. We write $EF$ for the event that a typical link $EF$ is present, and $EF^c$ for its complement. We write $A \leftrightarrow D$ for the event that $A$ is connected to $D$ by present links.

(d) $P(A \leftrightarrow D | AD^c) = P(A \leftrightarrow D | AD^c \cap BC^c) p + P(A \leftrightarrow D | AD^c \cap BC)(1-p)= \{1- (1-(1-p)^2)^2\}p + (1-p^2)^2 (1-p)$

I have some difficulty in interpreting this solution. For example, I understand that $$P(A \leftrightarrow D | AD^c \cap BC^c) = P((AB \cap BD) \cup (AC \cap CD)| AD^c \cap BC^c)$$

but how can this be equal to $ \{1- (1-(1-p)^2)^2\}$? (I guess that I need to use the complement several times to reach this, but I cannot figure out by myself.)

Any help would be appreciated.

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The probability that $AB$ and $BD$ are both present is $(1-p)^2$, so the probability that at least one of these links is absent is $1-(1-p)^2$; this is the probability that there is no connection between $A$ and $D$ via $B$. Similarly, the probability that at least one of the links $AC$ and $CD$ is absent, so that there is no connection between $A$ and $D$ via $C$, is $1-(1-p)^2$. The probability that at least one of the links $AB$ and $BD$ is absent and at least one of the links $AC$ and $CD$ is absent is therefore $$\big(1-(1-p)^2\big)^2\;.$$ Given that $AD$ and $BC$ are both absent, this is the probability that there is no connection between $A$ and $D$, since the only possible connections are via $B$ or $C$. Thus, $$1-\big(1-(1-p)^2\big)^2$$ is the probability that there is a connection between $A$ and $D$.