automobiles are randomly distributed with an average spacing of 1000 feet along a highway the probability that at least two cars are present in 1000 feet interval selected at random
Answers
Answer:
If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
Ad by OPPO Mobile India
Fully charged in 36 min - 65W SuperVOOC 2.0 in OPPO Reno4 Pro.
Sense the Infinite with superfast 65W SuperVOOC in OPPO Reno4 Pro. Charge for full day, just 36 min away!
Learn More
7 Answers
Nehal Amin, Mathematics and Computing,IIT Kgp
Answered June 1, 2018
For curiosity, I saw other solutions and found that everyone is solving via probability of not observing a car in 10 minutes.
Let's answer it in a straight way.
**Assumption**: Considering Probability of observing a car in any given non-overlapping time interval of equal length are equal and independent. **Reason:** Question clearly states "assuming constant default probability"
Let 'p' be the probability of observing a car in any 10 minutes interval.
Now let's generate the probability of observing a car in 30 minutes, let it be P(30).
Let's divide 30 minutes time interval into three 10 minutes int
Continue Reading
Related Questions (More Answers Below)
If the probability of seeing a cat on the highway is 0.95 in 30 minutes, what is the probability of seeing a cat on the highway in 10 minutes?
816 Views
The probability of seeing a car on the highway in 30 minutes is 0.95. What is the probability of seeing a car on the highway in ten minutes?
991 Views
The probability of a car passing is 0.95. What is the probability of a car passing in half an hour?
460 Views
On a biased die, an even number appears three times as frequently as an odd number. If the die is thrown twice, what is the probability that the sum is 9 or more?
3,790 Views
What is the probability of observing a car in 10 minutes (assuming constant default probability)?
470 Views
Other Answers
Tarun Malviya, former Derivatives Trader
Answered November 9, 2014 · Author has 88 answers and 160.5K answer views
Let say probability is of observing a car in 10 minutes be p, so not observing a car is 1-p
Now, when you say "observing a car" it means seeing at least one car.(This is what I believe meaning of the statement, as I think you stay on the road and wait for a car as soon as you see one you count that trial as positive and do other trials to arrive at value 0.95, this is my model for the problem)
So probabibilty that you see a car in 30 main is 1-(1-p)^3 (again 1-none gives you at least one)
This is given to be 0.95
So, 1-(1-p)^3=0.95
So, 1-p= 0.05^(1/3)
p=1-0.05^(1/3), the required probability
In fact if you divide 30 in n interval you will get 1-(0.05)^(1/n)
P.S. if you take "a car" as one car then you observe only one in 3 10 minute so you get 3*p(1-p)^2=0.95 solve this to get p which is about 1.45 so not possible.
Maximum value of 3p(1-p)^2 is 4/9 in (0,1) so unless you have value for 30 minutes less than this you won't get any solution.
Answer:
sorry dear I have given you irrelevant answer because I need some points
I hope you will understand
be my friend by following me
thank you