Two friends have decided to meet between 3 and 4 o clock and they have decided to wait for 10 mins to meet (for eg A have reached by 3 he will wait for 10 mins for B similarly B also ) then what is the probability of both are meeting?
Solution(By Examveda Team)
The time period within which they can arived is 1 hour. They have to be with 10 minitutes of each other.
The probability will vary 10/60 to 20/60 and it will remain constatnt 3:50 and then decrease to 10/60.
The required probability will be,
= 1 - unfavorable condition
= 1 - 2* [((20/60)*(30/60)*(40/60)*(50/60)*(50/60)]
= 1 - 2*[10/108]
= (108 - 20) /108
= 88/108 = 11/27
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formula is (w1+w2-(w1^2)/2-w2^2/2)
w1-waiting time of one person/totaltime(min)
w2-waiting time of another person/totaltime(min)
so
=w1=10/60,w2=10/60
=1/6+1/6-{(1/6)^2/2+(1/6)^2/2}
=2/6-{2/72}
=(24-2)/72
=22/72
=11/36