TWO PERSONS A AND B AGREE TO MEET AT A PLACE BTWEEN 11 TO 12 NOON. THE FIRST ONE TOARRIVE WAITS FOR 20 MIN AND THEN LEAVE. IF THE TIME OF THIR ARRIVAL BE INDEPENDET AND AT RNDOM,THEN THE PROBABILITY THAT A AND B MEET IS

[A]1/3 [B]2/3 [C]4/9 [D]5/9

time for meeting is 1 hr ie 60 min an dwaiting time is 20 min that means a person is present at that place for 1/3 of the time

we find the prob that they dont meet = (A comes x Bdoesnt) + (B comes A doesnt)

= 1/3x2/3 + 1/3+2/3 = 4/9

the prob that they meet is 1-4/9 = 5/9 

PLs do tell the answer to this question it is an interesting one

ANSWER IS [A]

AS THE FIRST PERSON TO ARRIVE WAITS FOR 20 MINUTES N(E)=20

TOTAL TIME IS 1 HOUR SO N(S)=60

THE OTHER PERSON HAVE TO ARRIVE IN THAT 20 MINUTES

THEREFORE PROBABILITY=20/60=1/3