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Let's say that a certain measure 'X' has a 60% chance of preventing an event and another measure 'Y' has a 70% chance of preventing the same event. Let's say that both measures are used, then what would be the odds of preventing that same event if the options are a.independent and b.dependent?

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    $\begingroup$ The answer is going to depend on whether those two options are independent or not. $\endgroup$ – dbx Oct 9 '17 at 16:12
  • $\begingroup$ Not enough information to answer. $\endgroup$ – JMoravitz Oct 9 '17 at 16:12
  • $\begingroup$ Now that you've added "if the options are independent" (which is not a safe assumption in many real world examples), we are looking for $Pr(X\cup Y)$, the probability that having applied measures X and Y to attempt to prevent the event that at least one of them was successful at preventing the event. We have then $Pr(X\cup Y)=Pr(X)+Pr(Y)-Pr(X\cap Y)=0.6+0.7-0.6\cdot0.7=1.3-0.42=0.88$. Note the use of the independence assumption in simplifying $Pr(X\cap Y)$ as $Pr(X)\cdot Pr(Y)$ which is otherwise not true. $\endgroup$ – JMoravitz Oct 9 '17 at 16:18
  • $\begingroup$ "Add probabilities". So 70% + 60% = 130%. Well..... $\endgroup$ – fleablood Oct 9 '17 at 16:34
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Another way of answering this question would be supposing the measures run one after the other and adding the probability of X preventing the event (60%) and the probability of Y preventing the event after X has failed to do so. That's 40% (probability of X not preventing the event) times 70% (probability of Y stopping the event).

This is $0.6 + 0.4\times0.7 = 0.6 + 0.28 = 0.88$

When you have many measures to consider, working with disjoint sets is, in my opinion, a lot easier

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If the events are independent then we use the familiar formula,

$P(X \cup Y)= P(X)+P(Y)-P(X \cap Y)$

$=0.6+0.7-(0.6)(0.7)=0.88$

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