# basic probability question?

• May 22nd 2013, 07:24 AM
jickjoker
basic probability question?
can anyone help with this question , I found it on a website cant work out how to get to the answer. On a deserted road, the probability of observing a car during a thirty-minute period is 95%. What is the chance of observing a car in a ten-minute period

the answer is given on the site , but cant work out how to get to it ! (Headbang)
• May 22nd 2013, 07:47 AM
ebaines
Re: basic probability question?
Let P(x) = probability of seeing at least one car in x minutes: $\small P(30)= 0.95$. The probability of seeing at least one car in 30 minutes is one minus the probability of not seeing a car in 30 minutes, or $\small P(30) = 1- \overline {P(30)}$. Next, $\small \overline {P(30)} = \overline {P(10)}^3$. In other words, to not see a car in 30 minutes is the same as not seeing a car in 10 minutes three times in a row. Hence:

$P(30) = 0.95 = 1 - \overline {P(10)}^3 = 1-(1-P(10))^3$

From this you solve for P(10), and get $\small P(10) = 0.6316$
• May 22nd 2013, 11:06 AM
jickjoker
Re: basic probability question?
hey ebaines thanks for your help , those formulas are far away from my understanding , ive never been to university im a basic calculator guy! but as regards to the question , its like going back in time , 95% chance in 30 min , as we get to 20 min, then 10 min , the chances of seeing a car get less and less , just struggling to do it on a basic calculator , il keep trying
on my calculator to see if I can get to your answer !
• May 22nd 2013, 11:57 AM
ebaines
Re: basic probability question?
One way to show that the anwer I gave is correct is to see if it leads to a 95% chance of seeing a car in 30 minutes. You can do the math like this: consider the thirty minutes as being cut up into 3 ten-minute sessions, call them sessions A, B and C. If you see one or more cars in 30 minutes it's because one of the following events occurred:

1. You see a car is session A, but not in B or C. The probability of this is 0.6316 x (1-0.6316)^2 = 0.08572
2. You see a car in session B, but not in A or C. The probability of this is again 0.08752
3. You see a car in session C. but not in A or B. Probability = 0.08752
4. You see a car in both sessions A and B, but not C. The probability of this occuring is (0.6316)^2 x (1-0.6316) = 0.14696
5. You see a car in both sesssions A and C but not B. The probabilit uof this is again 0.14696
6. You see a car in sessions B and C, but not A. The probability is 0.14696
7. You see a car in all three sessions A, B and C.The probability of this is (0.6316)^3 = 0.25196

If you add the probabilities of all seven possibilities it comes to 0.95, as expected.

Another way to double check is to consider that the probability of meeting a car in 30 minutes is equivalent to one minus the probability of NOT meeting a car in 30 minutes. The probability fo not meeting a car in 30 minutes is equal to the probability of not meeting a car in ten minutes cubed. So P(30) = 1 - P(Not 10)^3 = 1-0.3684^3 = 0.95. So again it checks.
• May 23rd 2013, 09:57 AM
jickjoker
Re: basic probability question?
hey ebaines I never doubted you was correct! I was just struggling with how to get from 095 for 30 minutes , to 0.6316 for 10 minutes . but after looking at your answer I realize the answer is from following the cube root of 5% , cube root of 0.05 =03684 minus 1 = 06316 , so if the question was for 20 minutes instead of 10 the answer would be 0.8642 ? I think that's correct?
• May 23rd 2013, 10:12 AM
ebaines
Re: basic probability question?
Quote:

Originally Posted by jickjoker
if the question was for 20 minutes instead of 10 the answer would be 0.8642 ? I think that's correct?

Yes - that's correct:

P(20) = 1 - P(not 20) = 1 - (P(not 10))^2 = 1 - (1-0.6316)^2 = 0.8642