Consider a post office with two clerks. Three people, A, B, and C, enter simultaneously. A and B go directly to the clerks, and C waits until either A or B leaves before he begins service. What is the probability that A is still in the post office after the other two have left when

A) 0

B) 1/27

C) 1/4

Step-by-step explanation:

Probability is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. The formular is represented below:

P(A) = Number of Favorable Outcomes / Total Number of Possible Outcomes

Let S1, S2, S3 represent the service times for A , B and C respectively

a) PS1 > S2 + S3 =0

Thus there is no probability that A is still in the office after the other two have left. The probability is 0.

b) Let

θ

θ be the possible values S1, S2 , S3 then

S1 > S2 + S3

⇔

⇔ S1 = 3 , S2 = 1 , S3 = 1

P( S1 > S2 + S3 ) = P(S1=3)P(S2=1)P(S3=1)

⇔

⇔ 1/3 x 1/3 x 1/3 = 1/27

(c) P(X >Y) = 1/2 where X,Y belongs { A, B , C}

where A,B,C represents the waiting times of clients A,B,C respectively. Now due to the memoryless property of exponential distribution. we have that

P(A>B+C)

=P(A>B+C/A>B)P(A>B)

=P(A>C)P(A>B)

=1/2X1/2

=1/4

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation: