Math Is Fun Forum

lindah

Member

Registered: 2010-01-25

Posts: 121

Probability - different time frames

Hi all,
May I verify if my thinking is spot on [I've exhausted the ideas, and this is the best I can come up with ]?

Thank you in advance for any assistance!

If the chances of seeing a meteor outside within an hour are 84%, what is the chance of seeing at least one new meteor outside in 30 minutes?

P(see meteor at least once within hr) = 1 - [P(seeing none in 30 min)]^2

= 0.6

and

If the probability of observing a car in 20 minutes is 609/625, What is the probability of observing at least one car in five minutes?

P(at least one car in 20 minutes) = 1 - [P(no cars in 5 minutes)]^4

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability - different time frames

Hi lindah;

You got me. What part of statistics are you studying that gave you that formula?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

qcao

Guest

Re: Probability - different time frames

I think your problem could have been stated more clearly. 
Let q be the probability of observing at least a car in 20 minutes, q = 609/625.
1-q = probability of observing no car in 20 minutes.
Let p be the probability of observing at least a car in 5 minutes.
1-p = probability of observing no car in 5 minutes.
Assume that p is the same for the four 5-minute invervals contained in 20 minutes. 
P(seeing no car in 20 minutes) = (1-p)^4.
1-q = (1-p)^4
1-p = (1-q)^(1/4)
p = 1 - (1-q)^(1/4) = 1 - (1 - 609/625)^(1/4)
p = 1 - 2/5 = 3/5