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June 21st, 2018, 10:57 PM   #1
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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.
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June 22nd, 2018, 10:18 AM   #2
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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*}$
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June 22nd, 2018, 05:54 PM   #3
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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.
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June 24th, 2018, 07:16 AM   #4
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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?
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June 24th, 2018, 10:22 AM   #5
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Quote:
Originally Posted by 2Clueless View Post
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.
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June 25th, 2018, 09:37 AM   #6
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Quote:
Originally Posted by 2Clueless View Post
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.
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June 25th, 2018, 11:05 AM   #7
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P[Event3 passes|Event1 or Event2 passes]P[Event1 or Event2 passes]
= .75 * ( .75 * .2 + .25 * .8 )
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June 25th, 2018, 01:11 PM   #8
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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 01:31 PM.
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June 26th, 2018, 12:45 AM   #9
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Quote:
Originally Posted by romsek View Post
$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

$P[\text{Event3 passes}] = (0.11)(0.6) = 0.066$
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June 26th, 2018, 05:52 PM   #10
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Thumbs up

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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