Part – A:
The Traffic Police department is concerned with the increase in Accidents on Shaheed-Millat Expressway.They are setting up a dedicated team to analyse the causes and propose a solution. Accidents can occur because of the sharp turns on road or unavailability of Pedistrian bridges. On the other hand, accidents can occur due to unsafe driving or human error. The observation was further breakdowned into 4 shifts to observe the impact of traffic flow, Morning (7:00am to 10:00am), Day (10:00am-3:00pm), Evening (3:00pm – 7:30pm) and Night (7:30pm – 7:00am). During the one month of observation, 170 minor and major accidents have occurred. The percentages of the accidents for the condition combinations are given in Table 5:
SHIFT | UNSAFE ROAD CONDITIONS | HUMAN ERROR |
MORNING | 6% | 28% |
DAY | 4% | 12% |
EVENING | 5% | 26% |
NIGHT | 3% | 16% |
If an accident report is selected randomly from the 170 reports,
Part – B:
An University wishes to hire 4 faculty members for its newly established department. They have to select from a group of 8 Male and 7 Female applicants, all of whom are equally qualified to fill in the positions. If University selects 4 at random, what is the probability that ;
Part – C:
If the license plates in a Karachi consist of three letters followed by three numbers, what is the probability that a randomly generated plate ends in the numbers “786”? Consider that repetitions are allowed and order matters.
Part – D:
A students is going to graduate from a business department of a university by the end of the semester. After being interviewed at two companies he likes, he assesses that his probability of getting an offer from company A is 0.6, and his probability of getting an offer from company B is 0.45. If he believes that the probability that he will get offers from both companies is 0.35, what is the probability that he will get at least one offer from these two companies?
A)
a) P(evening) = 31%
b) P(human error) = 82%
c) P(unsafe road) = 18%
d) P(either the evening or the night shift) = 31%+19%=
50 %
B)
a) P(all men)=8C4/15C4= 0.0513
b) P(all women)=7C4/15C4= 0.0256
c) P(at least one women) = 1 - P(no women) =
0.9487
d) P(at least one men) = 1 - P(no men) =
0.9744
C)
required probability = 26^3*1/(26^3*10^3)=
0.001
D) P(at least one) = P(A) + P(B) - P(both) = 0.6+0.45-0.35=0.7
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