Answer: Probability That Mail Will Arrive Before Noon at Least One Day Wed/Thurs

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The discussion focuses on calculating the probability that mail will arrive before noon at least one day between Wednesday and Thursday, given an 80% chance for Wednesday and a 40% chance for Thursday. Participants clarify that these probabilities are independent, and the correct approach involves calculating the probability that mail does not arrive before noon on either day. By determining the probability of no delivery on both days as 0.2 (for Wednesday) multiplied by 0.6 (for Thursday), the result is 0.12. Subtracting this from 1 gives a probability of 0.88 or 88% for at least one day. The conclusion raises a question about whether this probability logically exceeds the individual day probabilities.
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There is a 80% chance that Wednesday's mail will be delivered before noon on Wednesday and a 40% chance that Thursday's mail will be delivered before noon on Thursday. If these probabilities are independent, what is the probability that the mail will arrive before noon at least one of those two days?
 
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What thoughts have you had so far?
 
heh guess I should be posting what I've done on these instead of just posting the question :)

ok well all I know is that its .8 (x) + .4 (x) , but I have no idea what the variable x is. I can't just add .8 and .4 because that would be greater then one. So i guess I just need help figuring out what x is
 
Well, why do you think that?
 
or maybe this has something to do with the independent test...
 
One way to go about solving "at least one" problems, is to think of the probability that none happen.
 
so 1-.2(.6)? which is 88% then?
 
Evilpony,

Does that answer (88%) seem to make sense? Should it be greater than the probability on either individual day?
 

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