# Can we add probabilities? [closed]

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?

• The answer is going to depend on whether those two options are independent or not. – dbx Oct 9 '17 at 16:12
• Not enough information to answer. – JMoravitz Oct 9 '17 at 16:12
• 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. – JMoravitz Oct 9 '17 at 16:18
• "Add probabilities". So 70% + 60% = 130%. Well..... – fleablood Oct 9 '17 at 16:34

This is $0.6 + 0.4\times0.7 = 0.6 + 0.28 = 0.88$
$P(X \cup Y)= P(X)+P(Y)-P(X \cap Y)$
$=0.6+0.7-(0.6)(0.7)=0.88$