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I have question on probability :

  • $H$ occurs $\implies$ $C$ occurs

  • $H$ does not occur $\implies$ $C$ can occur or cannot

  • if i want to compute $P(H|C)=\dfrac{P(H).P(C|H)}{P(C)}$ does $P(C|H)=1$ ? If yes then $P(H|C)=\dfrac{P(H)}{P(H)+P(H^c)(1-P(C))}$ ?
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    $\begingroup$ It seems strange to call this a "probability" problem! The only possible probabilities are "0" and "1"- "false" and "true" so I would consider this a logic problem. "If H is true then C is true is the same as "if C is false the H is false". Similarly "If C is true then H is true" is the same as "If H if false the C is false". The two together give "H is true if and only if C is true" Both P(H|C) and P(C|H) are 1. $\endgroup$
    – user247327
    May 19, 2017 at 18:31
  • $\begingroup$ If i say C can occur even if H was false would that be a different question? $\endgroup$
    – Keller
    May 19, 2017 at 18:40

2 Answers 2

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If we interpret $H \implies C$ as $H \subseteq C$, then $H \cap C = H$ so $$P(H \vert C) = \frac{P(H \cap C)}{P(C)} = \frac{P(H)}{P(C)} \le 1$$ while $$P(C \vert H) = \frac{P(H \cap C)}{P(H)} = \frac{P(H)}{P(H)} = 1$$

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  • $\begingroup$ yes this is it i think thank you that implie sign mean that H is in C $\endgroup$
    – Keller
    May 19, 2017 at 19:07
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In answer to the first question, Yes, $P(C|H)=1$.

In answer to the second question, No, $P(C)\not=P(H)+P(H^c)(1-P(C))$ in general. For example, suppose you roll a die twice. Let $H$ be the event that the sum of the two rolls is $11$, and let $C$ be the event that the first roll is a $5$ or a $6$. It's clear that if $H$ occurs then $C$ occurs (how can the sum be $11$ if the first roll is $4$ or less?). And it's easy to see that $P(H)={1\over18}$ and $P(C)={1\over3}$, and then check that

$$P(H)+P(H^c)(1-P(C))={1\over18}+{17\over18}(1-{1\over3})={37\over54}\not={1\over3}=P(C)$$

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