There are two trains, arriving between 8 to 8:20. First train hault for 3 min. and second halt for 4 min.
What is the probablity that second train will arrive before first train?
2007-02-04 20:55:28 · 4 answers · asked by SAM 1 in Science & Mathematics ➔ Mathematics
8:00 and 8:20 ,,,,,,8:20- 8:00 = 20 min
p1 = 3/20 and p2=4/20
p= 4/20 / (3/20 + 4 /20)
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2007-02-04 21:46:01 · answer #1 · answered by Anonymous · 0⤊ 0⤋
you do mean two independent tracks ?
train a hault for 3 min at the station you are observing.
train b hault for 4 min at the station you are observing.
what is the prob that b arrive before a leaves (given that a arrives first) ?
E[a] = mid point of 0 to 20 = 10 (8:00 to 8:20 am)
E[b] = mid point of 0 to 20 = 10.
find P[ b < a +3 | a arrives first ] = prob b arrives in 3 min after a arrived.
prob ( b arrives <= 3 mins) = 3/20.
2007-02-05 05:02:55 · answer #2 · answered by e_kueh 2 · 0⤊ 0⤋
If you mean "leaves" before the first train, that just means that the one who halts for 4 mns arrived more than one minute before the other. So this probability is
(19)^2/ 2x(20)^2 that's 361/800.
2007-02-04 22:01:08 · answer #3 · answered by gianlino 7 · 0⤊ 0⤋
0. The second train to arrive is always the 2nd train to arrive.
2007-02-04 21:00:04 · answer #4 · answered by Helmut 7 · 1⤊ 0⤋