Latest Tech Mahindra Aptitude Question SOLUTION: A and B decide to meet between 5 pm and 7 pm. It was decided that each will wait for the other for 20 mins and if the other person didnt come, he would leave. Find probability of t

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(#M40037499) TECH MAHINDRA Logical Reasoning question Probability Keep an EYE

A and B decide to meet between 5 pm and 7 pm. It was decided that each will wait for the other for 20 mins and if the other person didnt come, he would leave. Find probability of their meeting.

Asked In Tech Mahindra Debojit Pal (9 years ago)

Unsolved Read Solution (13)

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Read Solution (13) :

shruti saumya  (9 years ago)

total time allotted = 2*60 = 120 minutes.
time during which they can meet= 20 minutes
so, the probability= 20/120 = 1/6

Is this solution Helpfull? Yes (45) | No (9)  

Debojit Pal  (9 years ago)

Incorrect answer MAYANK SHARM,
This is an example of continuous function as there is no gap between time intervals.
Consider a graph with A on X axis and B on Y axis with origin at 5 pm
The end points of each A and B is 2 hrs.
The possibility of their meeting is given by the area between both of them
Total area=(2 hrs)^2=120^2
Region of both A and B is a triangle of base and height equal to 120-20=100
Reqd. probability=(120^2 - 100^2)/14400 = 11/36 is the required answer..

Is this solution Helpfull? Yes (19) | No (13)  

abhishek srivastava  (9 years ago)

hii shruti this is abhishek...tmhari aur meri soch bht milti hai(ans same hai na ummmmmm)...meet me at ..thapa momos.....

Is this solution Helpfull? Yes (13) | No (25)  

Sayantani Banerjee   (7 years ago)

Meeting occurs if the first person arrives between 5:00 and 6:40 and the second
person arrives in the next 20 minutes after the 1st friend arrives or if both the persons arrive between 6:40 and 7:00.
Case 1:
80/120 are the 1st probability
and hence probability of
first person arriving
between 5:00 and 6:40 is
5/6.
Probability of second person arriving in the next 20 minutes = 20/120 = 1/6
person arriving between
5:00 and 6:40 and meeting
the other = 5/6 * 1/6*2 =
5/18( 2 for choosing the first
arriving friend any one of the two)
Case 2:
Both friend
between 6:40and 7:00. So,
probability = 1/6*1/6=1/36
So, probability of a meet = 5/18
+ 1/36 = 11/36

Is this solution Helpfull? Yes (3) | No   

khushboo goenka  (9 years ago)

1/3
probability is 20+20/120

Is this solution Helpfull? Yes (2) | No (1)  

ramesh  (9 years ago)

11/ 36 is the correct answer

Is this solution Helpfull? Yes (1) | No   

Mayank Sharm  (9 years ago)

This can be solved exactly with the same concept of 'Minimum Length Problem'.

The time period in which they can arrive is 1 hour and they have to be within 20 minutes of each other or 0.333 units of each other.

=> The distance from the line A = B should be less than 0.25 units

Required probability = 1 – Unfavorable probability

= 1 – (Sum of the areas of the two triangles / Total Area)

= 1 – 2*(1/2 * 2/3 * 2/3) = 5/9

Please let me know if i am right

Is this solution Helpfull? Yes | No (7)  

Divyasri  (9 years ago)

if we want to solve this problem, we calculate for every minute.
It is impossible
they are just confusing..
To this qstn just think in a simple way.
So,probability of their meeting is just 1/2

Is this solution Helpfull? Yes | No (1)  

Tushar Sharma  (9 years ago)

4*(1/6)*(1/3)+2*(1/6)*(1/6)=1/2

Is this solution Helpfull? Yes | No (1)  

Sourav ganguly  (9 years ago)

considering probability curve the max chance of meeting will fall in between the coordinate (100,120) & (120,100) because 120-20=100 min will fall on common region

so the required probability will be (100*100)/(120*120)

Is this solution Helpfull? Yes | No   

ranjith reddy  (9 years ago)

1-prob of not meet
=1-((40/60)*(40*60))
=1-(16/36)
5/9
ans 5/9

Is this solution Helpfull? Yes | No (2)  

suman  (9 years ago)

b+b=5,tfghghh
ggyhh
hhhjjhj
so b=5

Is this solution Helpfull? Yes | No   

aman jain  (7 years ago)

0.3298 using graph

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