You arrive at a bus stop at 10 o’clock, knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30. (a) What is the probability that you will have to wait longer than 10 minutes? (b) If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?

A) 2/3

B) 1/3

Step-by-step explanation:

A) Since the bus arrives uniformly at the bus stop between 10 and 10:30.Let X denote the arrival of a bus at a bus stop. Thus X is a uniform tandom variable having a density function;

f(x) = 1/30 : x ∈ [0,30]

Now, probability that a passenger arriving at the bus stop between 10 and 10:30 will have to wait more than 10 minutes can be expressed as;

P[x > 10] = ∫f(x)•dx at boundary of 30 and 10

Thus,

∫(1/30)•dx at boundary of 30 and 10

P[x > 10] = (1/30)•x at boundary of 30 and 10

P[x > 10] = (1/30)(30) - (1/30)(10)

P[x > 10] = 1 - 1/3 = 2/3

B) The probability that the passenger will have to wait additional 10 minutes is given as;

P[X > 25 | X > 15] =P[X > 25 ∩ X > 15]

= (P[X > 25])/(P[X > 15])

= [∫(1/30)•dx at boundary of 30 and 25]/ [∫(1/30)•dx at boundary of 30 and 15]

= [(1/30)•x at boundary of 30 and 25]/[(1/30)•x at boundary of 30 and 15]

= [(30/30) - (25/30)]/[(30/30) - (15/30)]

= (1 - 5/6)/(1 - 1/2)

= 1/6 ÷ 1/2 = 1/3

∩

the correct answer is the second option