My Math Forum Another Probability Question
 User Name Remember Me? Password

 Probability and Statistics Basic Probability and Statistics Math Forum

 June 21st, 2018, 09:57 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 10:13 PM.
 June 22nd, 2018, 09:18 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 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 - (1-0.25)(1-0.2)(1-0.11) = \\ \\ &1 - (0.75)(0.8)(0.89) = 0.466 \end{align*}
 June 22nd, 2018, 04:54 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, 06:16 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, 09:22 AM   #5
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,530
Thanks: 1390

Quote:
 Originally Posted by 2Clueless 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?
How about you show me what you've done so far.

June 25th, 2018, 08:37 AM   #6
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,530
Thanks: 1390

Quote:
 Originally Posted by 2Clueless 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?
$P[\text{Event3 passes}] = P[\text{Event3 passes|neither Event1 or Event2 passes}]P[\text{neither Event1 or Event2 passes}] + P[\text{Event3 passes|Event1 or Event2 passes}]P[\text{Event1 or Event2 passes}]$

This can clearly be reduced to

$P[\text{Event3 passes}] = P[\text{Event3 passes|neither Event1 or Event2 passes}]P[\text{neither Event1 or Event2 passes}]$

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, 10:05 AM #7 Newbie   Joined: Jun 2018 From: Toronto Posts: 2 Thanks: 0 P[Event3 passes|Event1 or Event2 passes]P[Event1 or Event2 passes] = .75 * ( .75 * .2 + .25 * .8 )
June 25th, 2018, 12:11 PM   #8
Newbie

Joined: Jun 2018
From: USA

Posts: 6
Thanks: 0

Thank you all for the replies.

mrinmayk
Quote:
 P[Event3 passes|Event1 or Event2 passes]P[Event1 or Event2 passes] = .75 * ( .75 * .2 + .25 * .8 )
If I did the arithmetic right, this = 0.2655 or approximately 26%?

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 12:31 PM.

June 25th, 2018, 11:45 PM   #9
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,530
Thanks: 1390

Quote:
 Originally Posted by romsek $P[\text{Event3 passes}] = P[\text{Event3 passes|neither Event1 or Event2 passes}]P[\text{neither Event1 or Event2 passes}] + P[\text{Event3 passes|Event1 or Event2 passes}]P[\text{Event1 or Event2 passes}]$ This can clearly be reduced to $P[\text{Event3 passes}] = P[\text{Event3 passes|neither Event1 or Event2 passes}]P[\text{neither Event1 or Event2 passes}]$ 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.
\begin{align*} &P[\text{neither Event1 or Event2 passes}] =\\ \\ &P[\text{Event1 fails}]P[\text{Event2 fails}] = \\ \\ &(1-0.25)(1-0.2) = \\ \\ &(0.75)(0.8) = 0.6 \end{align*}

So

$P[\text{Event3 passes}] = (0.11)(0.6) = 0.066$

 June 26th, 2018, 04:52 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!

 Tags probability, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post HipHopper Advanced Statistics 1 April 18th, 2013 07:28 PM Chee Probability and Statistics 1 April 14th, 2012 04:53 AM edvinn Probability and Statistics 1 October 15th, 2011 08:33 AM hoyy1kolko Advanced Statistics 1 May 31st, 2011 11:56 PM djbip Probability and Statistics 11 June 15th, 2010 09:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.