Show that if $P(C) = 1$, then $P(D|C) = P(D)$ using probability rules.

I tried using conditional probability to get from $P(D|C) = P(D \cap C) / P (C) = P(D \cap C)$ but I am not sure how to continue the proof to get to $P(D)$. Can someone explain what the next steps I should do in order to get to $P(D)$?


I would solve it like this:

$$ \begin{aligned} P(D|C) &= P(D\cap C) / P(C) \\ &= P(D \cap C) \\ &= P(D) + P(C) - P(D \cup C) \\ &= P(D) + 1 - 1 \\ &= P(D) \end{aligned} $$

Where our third equation is explained by the rule of addition with some simple rearrangement of terms:

$$ P(D \cup C) = P(D) + P(C) - P(D \cap C)$$ $$ \therefore P(D \cap C) = P(D) + P(C) - P(D \cup C)$$

And our fourth equation is explained by the fact that $P(D \cup C) = 1$, since it is the probability of either $D$ or $C$ occurring, and we know that $P(C) = 1$

  • $\begingroup$ Very neatly done. $\endgroup$ – user140401 Dec 5 '16 at 10:49

I would approach it through a frequentist definition of probability, by showing that the set of events that count as D ∩ C has the same count as the set of events that count as D. That is P(D ∩ C) = N(D ∩ C)/N(total) and P(D) = N(D)/N(total) can be shown to be the same if every event D is also an event D ∩ C (so that N(D ∩ C) is the same as N(D)). Because C is universally true, this will be the case.

Edit: Okay - a more extended answer based on your comment

P(D|C) = P(D ∩ C)/P(C)
P(D|C) = P(C|D)*P(D)/P(C)

so we have to show that P(C|D) = 1, as does P(C), which was given.

P(C) = P(C|D)P(D) + P(C|!D)(1-P(D))

Let P(C|D) = 1-a and P(C|!D) = 1-b.
a, b must be nonnegative by the rules of probability (0-1 range)
1 = (1-a)P(D) + (1-b)(1-P(D))
1 = P(D) - a
P(D) + 1 - P(D) - b + b*P(d)
b(1-P(D)) = -aP(D)

which, given the 0-1 constraints on a,b,P(D) is only possible if i) both a and b are zero, ii) both b and P(D) are zero, or iii) a=0 and P(D) = 1

In the first case and third case, we have that P(C|D) = 1 because a=0 and we have proved what was required.

In the second case, if P(D) = 0 then we can show that this can only be satisified when P(D|C) is zero since P(D) = P(D|C)*P(C)+.... in which case we have still proved that P(D) = P(D|C) as required (since they both = 0).

  • $\begingroup$ Yes the above seems to work, but I need to solve this question using probability rules without using frequentist definition of probability. If you know how should I proceed that way, let me know. $\endgroup$ – user141156 Dec 5 '16 at 5:23
  • $\begingroup$ Since the very meaning of probability, as captured in its axioms, is separate from one's interpretation or philosophy of probability, any argument from a "frequentist definition"--at least if it is rigorous--would be identical to an argument from any other point of view. $\endgroup$ – whuber Mar 16 '18 at 16:57

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