Question

Probability of a truck getting past of a car in 1 hr is 0.2 What is the probability of the truck getting past the car in 2 hrs!?

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Answer 1

Step-by-step explanation:

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)?

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 intervals as A, B and C.

P(A) = Probability of seeing a car in first 10 minutes

P(B) = Probability of seeing a car in second 10 minutes

P(C) = Probability of seeing a car in third 10 minutes

As all are independent events so,

P(A) = P(B) = P(C) = p

Similarly,

P(not A) = Probability of not seeing a car in first 10 minutes

P(not B) = Probability of not seeing a car in second 10 minutes

P(not C) = Probability of not seeing a car in third 10 minutes

As all are independent events so,

P(not A) = P(not B) = P(not C) = 1-p

Then,

P(30) = P(A) + P(not A)*P(B) + P(not A)*P(not B)*P(C)

It can be seen in this way:

Consider an event, We are tossing a coin 3 times a row and we want to find what is the Probability of getting at least 1 head.

P(getting at least 1 head) = P(getting head in 1st toss) + P(getting head in 2nd toss given in 1st toss we got tail) + P(getting head in 3rd toss, given in 1st and 2nd toss we got a tail)

P(getting at least 1 head) = 1/2 + 1/2*1/2 + 1/2*1/2*1/2 = 7/8

Similarly,

P(30) = p + (1-p)*p + (1-p)*(1-p)*p

=> 0.95 = p + p - p^2 + p + p^3 - 2p^2

=> 0.95 = p^3 - 3p^2 + 3p

=> 1-0.95 = 1 - p^3 + 3p^2 - 3p

=> 0.05 = (1-p)^3

=> p = 1 - (0.05)^(1/3)

=> p ~= 0.6316

P(probability of observing a car in 10 minutes) = 0.6316

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 2

Step-by-step explanation:

dii one more question please help me