A person goes to office either by car, scooter, bus or train probability of which being 1/7, 3/7, 2/7 and 1/7 respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, then what is the probability that he travelled by train ?
2007-01-23 21:20:34 · 7 answers · asked by Saumya Gupta 1 in Science & Mathematics ➔ Mathematics
This problem requires the use of Bayesian analysis.
p(c,s,b,t) = {1/7,3/7,2/7,1/7}
p(late | c,s,b,t) = {2/9,1/9,4/9,1/9}
p(on time | c,s,b,t) = {7/9,8/9,5/9,8/9}
Find p(train | on time).
p(train | on time) = p(train ∩ on time)/p(on time)
= (1/7)(8/9) / {(1/7)(7/9) + (3/7)(8/9) + (2/7)(5/9) + (1/7)(8/9)}
= 8/(7 + 24 + 10 + 8) = 8/49
2007-01-23 21:44:46 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
The probabilty that he goes by car and arrives on time is 1/7 x 7/9=7/63, for scooter 3/7x8/9= 24/63, bus 2/7x5/9=10/63 and train
1/7x8/9=8/63. Total probability of being on time is (7+24+10+8)/63=
49/63. Probability of train being taken is 8/63 / 49/63 = 8/49.
However the question makes no sense because since he already reached office, there is no randomness involved. He took the train with probability 0 or 1 depending on whether he took it or not...
2007-01-24 05:35:12 · answer #2 · answered by gianlino 7 · 0⤊ 0⤋
P(reached on time and travelled by train)=1/7 x 8/9
=8/63
2007-01-24 05:28:18 · answer #3 · answered by A 150 Days Of Flood 4 · 0⤊ 0⤋
1/7
2007-01-24 05:29:02 · answer #4 · answered by Mr. Dev 2 · 0⤊ 0⤋
using the the BAYE's theorem,
the answer is = (8/9) * (1/7) divided by whole in bracket
( (8/9* 1/7)(5/9 * 2/7)(8/9 * 3/7)(7/9 * 1/7) )
u calculate the answer now.
2007-01-24 05:35:40 · answer #5 · answered by krishna 4 · 0⤊ 0⤋
ask that man
2007-01-24 05:27:10 · answer #6 · answered by n nitant 3 · 1⤊ 2⤋
i like burgars............and pizza too
mmm....arent there any cartoons yet???!!!
2007-01-24 05:55:21 · answer #7 · answered by Anonymous · 0⤊ 1⤋