See a car
Problem:
If the probability of observing a car on a highway in 20 minutes time is 609/625 then what is the probability of observing a car in 5 minutes time on the same highway (considering all the factors involved to be uniform)?
Solution:
Highlight the part between the * symbols for the answer.
*Probability of seeing a car in 20 minutes = 609/625
Hence, the Probability of seeing a car in 5 minutes = 1 - 2/5 = 3/5*
If the probability of observing a car on a highway in 20 minutes time is 609/625 then what is the probability of observing a car in 5 minutes time on the same highway (considering all the factors involved to be uniform)?
Solution:
Highlight the part between the * symbols for the answer.
*Probability of seeing a car in 20 minutes = 609/625
=> Probability of not seeing a car in 20 minutes = 1 - 609/625 = 16/625
=> (Probability of not seeing a car in 5 minutes)^4 = 16/625
=> Probability of not seeing a car in 5 minutes = (16/625)^(1/4)
=> Probability of not seeing a car in 5 minutes = 2/5
23/25
ReplyDeleteNo, It will be 3/5.
ReplyDeleteIdea:- Prob that no event in two half intervals = prob that there is no event in full interval.
Therefore, 1-(1-p)^2=p' and 1-(1-p')^2=609/625. Where p is no event in the half intervals of first 10 minutes and p' is no event in the half interval of 20 minutes.
Therefore, p'=21/25 and p=3/5.
It will be 3/5.
ReplyDeleteIdea:- Prob that no event in two half intervals = prob that there is no event in full interval.
Therefore, 1-(1-p)^2=p' and 1-(1-p')^2=609/625. Where p is no event in the half intervals of first 10 minutes and p' is no event in the half interval of 20 minutes.
Therefore, p'=21/25 and p=3/5.