From where he stands, one step toward the cliff would send a drunken man over the edge. He takes random steps, either toward or away from the cliff. At any step his probability of taking a step away is 0.75, of a step toward the cliff is 0.25. What is his probability of escaping the cliff?
2007-06-12 04:53:05 · 4 answers · asked by prash 1 in Science & Mathematics ➔ Mathematics
0.733333 (1- (0.25 + 0.25^3 + 0.25^5 + 0.25^7 + 0.25^9 + ... + 0.25^(2n+1)))
2007-06-12 05:00:43 · answer #1 · answered by yeeeehaw 5 · 0⤊ 0⤋
You have to define the number of steps that is involved. For example, he may take 1 step backwards (away from cliff) and then take 2 steps forward thus falling over. Or he can take 1 step backward, 1 step forward, 2 steps backward and 3 steps forward etc.
The probability is defined by the binomial function (p + q)^n where p is the probability of falling and q is the probability of not falling and n is the number of trials.
2007-06-12 05:00:01 · answer #2 · answered by Swamy 7 · 0⤊ 1⤋
Since you've not mentioned how many steps are required to escape the cliff, lets assume this one step is the decider. In that case, the probability that he will escape is the probability of him stepping away, i.e, .75
2007-06-12 04:57:23 · answer #3 · answered by Nikhil M 3 · 0⤊ 1⤋
3/4
2007-06-12 04:56:37 · answer #4 · answered by Anonymous · 0⤊ 1⤋