To cross a single lane of moving traffic, we require at least a duration d. successive car interarrival times are independently and identically distributed with probability density function ft (t). if an interval between successive cars is longer than d, we say that the interval represents a single opportunity to cross the lane. assume that car lengths are small relative to intercar spacing and that our experiment begins the instant after the zeroth car goes by. determine, in as simple form as possible, expressions for the probability that: (a) we can cross for the first time just before the nth car goes by. (b) we shall have had exactly m opportunities by the instant the nth car goes by. (c) the occurrence of the mth opportunity is immediately followed by the arrival of the nth car.

the correct answer is option c.   3

step-by-step explanation:

points to remember

equation of the line with coordinates (x₁, y₁) and (x₂, y₂) is given by,

slope m = (y₂ - y₁)/(x₂ - x₁)

from the figure we can see a line with two coordinates given

to find the slope of given line

here   (x₁, y₁)   = (-1, -1 ) and (x₂, y₂) = (1, 5)

slope m = (y₂ - y₁)/(x₂ - x₁)

  = (5 - -1)/(1 - -1)

  = (5 + 1)/(1 + 1)

  = 6/2 = 3

therefore the correct answer is option c.   3

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for

correct answer is b. 27/8

step-by-step explanation: