Microsoft Interview Question for Senior Software Development Engineers


Country: India
Interview Type: In-Person




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11
of 11 vote

Let the probabilty of passing of bus in the time intersection of 5 min be X.
Then the probabilty of not passing of bus in the same 5 minute intersection is Y=1-X.

Let the bus not pass for 4 '5 min intersection', i.e, for 20 mins.
The probablity of above will be Y^4.

Then as per question,

1 - Y^4 = 0.9
Y^4 = 0.1
(1-X)^4 = 0.1
X = 0.4377

- aasshishh April 27, 2013 | Flag Reply
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0
of 0 votes

Shouldn't it be

(1-Y)^4 = 0.1

?

(1-Y)^4

is the probability that the bus would NOT pass for 20 mins.
0.9 is the probability that the bus WOULD pass in 20 mins.

- whatever September 26, 2014 | Flag
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0
of 0 votes

Nevermind, I made a misread Y as X.

You do say

(1-X)^4 = 0.1

- whatever September 26, 2014 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

The first solution is correct if the bus can only arrive once during the 20 minute period. The second is correct if the bus can arrive once during each 5 minute time slot.

The problem as stated is incomplete and requires further information for a definitive solution to be given.

- Both solutions are correct April 27, 2013 | Flag Reply
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0
of 0 vote

A continuous probability question. Divide 20 min of time in 5 min time slots. So 4 such time slots. Bus may pass through intersection in any one of the 4 slots. So probability that the bus passes through intersection in first 5 min is 0.9*1/4 = 0.225.

- lyf.pboy April 27, 2013 | Flag Reply
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0
of 0 vote

Some body can explain why 0.4377 is correct?

- anil2878 April 28, 2013 | Flag Reply
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0
of 0 vote

Assuming arrival of buses is a poisson process, the interarrival time of buses T ~ exp (lambda min).. lambda unknown.
Now Given P (T <= 20 min) = 1 - exp (-lambda * 20) = 0.9.
=> lambda = 0.115 min
=> P (T <= 5 min) = 1 - exp (-lambda * 5) = 0.43729

- chayan April 30, 2013 | Flag Reply
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0
of 0 vote

Let X be the time taken by bus to come out from intersection. Hence, it will be uniformally distributed in interval (0,b).
P(X <= 20)=0.9 which implied b= 20/0.9. Therefore, P(X<= 5)= 0.9*5/20=0.225.

- samar October 28, 2013 | Flag Reply
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0
of 0 vote

Let,
p(5) : probability of seeing no bus in 5 mins.
P(10): p(5)*p(5) = p(5)^2
similarly, p(20) = p(5)^4
also,
1-p(20) = probability of seeing bus in 20 mins
1-p(20) = 0.9 ....(given in question)
p(20)= 0.1
but, p(20) = p(5)^4
hence p(5) = p(20)^1/4
p(5) = 0.1^1/4 = 0.5623
p(seeing a bus in 5 mins) = 1- p(seeing no car in 5 mins)
p(seeing a bus in 5 mins) = 1-0.5623
p(seeing a bus in 5 mins) = 0.4377

Therefore probability of seeing a bus is 0.4377

- Yash Mehta (India) December 24, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

- Anonymous July 02, 2013 | Flag Reply


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