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June 21st, 2018, 10:57 PM  #1 
Newbie Joined: Jun 2018 From: USA Posts: 6 Thanks: 0  Another Probability Question
I hope this is an appropriate forum for this question. Apologies if it is not. Suppose there are 3 events, each of which will have either a "Pass" or "Fail" outcome. The probabilities of a "Pass" outcome for each event is: Event1 25% Event2 20% Event3 11% What is the probability that at least 1 of the 3 events will have a "Pass" outcome and how is it calculated? Thank you for any help! Last edited by 2Clueless; June 21st, 2018 at 11:13 PM. 
June 22nd, 2018, 10:18 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224 
Without further information we have to assume the events are independent. Probably the easiest way to go about this is $\begin{align*} &P[\text{at least 1 pass}] = \\ \\ &1  P[\text{0 passes}] = \\ \\ &1  (10.25)(10.2)(10.11) = \\ \\ &1  (0.75)(0.8)(0.89) = 0.466 \end{align*}$ 
June 22nd, 2018, 05:54 PM  #3 
Newbie Joined: Jun 2018 From: USA Posts: 6 Thanks: 0 
Thank you romsek! Yes, the assumption is correct. Each event is independent. I had guessed the probability was about 50% just by intuition. Pretty close! Thank you. This is very helpful to know. 
June 24th, 2018, 07:16 AM  #4 
Newbie Joined: Jun 2018 From: USA Posts: 6 Thanks: 0 
Okay... Question #2 Suppose now that the events are not independent. If Event1 *or* Event2 Pass, then the probability that Event3 will Pass equals 0%. What is the probability that Event3 will Pass? 
June 24th, 2018, 10:22 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224  
June 25th, 2018, 09:37 AM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224  Quote:
This can clearly be reduced to $P[\text{Event3 passes}] = P[\text{Event3 passesneither Event1 or Event2 passes}]P[\text{neither Event1 or Event2 passes}] $ You don't specify it but I'll assume that the first probability in that product is the same $P[\text{Event3 passing}]=0.11$ You can work out the second probability in that product as was done in the first question.  
June 25th, 2018, 11:05 AM  #7 
Newbie Joined: Jun 2018 From: Toronto Posts: 2 Thanks: 0 
P[Event3 passesEvent1 or Event2 passes]P[Event1 or Event2 passes] = .75 * ( .75 * .2 + .25 * .8 ) 
June 25th, 2018, 01:11 PM  #8  
Newbie Joined: Jun 2018 From: USA Posts: 6 Thanks: 0 
Thank you all for the replies. mrinmayk Quote:
But that seems intuitively strange to me. If the probability of Event3 passing independent of the other Events is 11%, and the other events are then linked to Event3 such that they might cause Event3 to have a probability of 0, depending on their outcome, shouldn't that reduce the probability that [Event3 passes] below 11%? Sorry if I misinterpreted or misstated my question. It's entirely possible since my skills are abominable. romsek Thank you. Your knowledge of this topic is impressive although I am not sure I even have enough knowledge of it to complete the calculations from your equation. I am 40 years out of college. I'm afraid my math skills are reduced to vapor. :eek Last edited by 2Clueless; June 25th, 2018 at 01:31 PM.  
June 26th, 2018, 12:45 AM  #9  
Senior Member Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224  Quote:
&P[\text{neither Event1 or Event2 passes}] =\\ \\ &P[\text{Event1 fails}]P[\text{Event2 fails}] = \\ \\ &(10.25)(10.2) = \\ \\ &(0.75)(0.8) = 0.6 \end{align*}$ So $P[\text{Event3 passes}] = (0.11)(0.6) = 0.066$  
June 26th, 2018, 05:52 PM  #10 
Newbie Joined: Jun 2018 From: USA Posts: 6 Thanks: 0 
Thank you romsek! Beautifully done! And in the range I intuitively expected. Also, after studying it, I understand your calculation although it's doubtful I could have constructed that equation from scratch. How I admire your skills! Use them or lose them. I'm afraid mine have been long lost! Thank you all for your help! 

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