A retired person chooses randomly one of the six parks of his town everyday and goes there for hiking. We are told that he was seen in one of these parks, Oregon Ridge, once during the last 10 days. WHat is the probability that during this period he has hiked in this park two or more times?
the answer should be 0.615, how to get that?I always got the wrong answer. thanks guys!
maths-probability
2007-01-31 2:24 am
回答 (2)
2007-01-31 3:07 am
✔ 最佳答案
P(he has hiked in this park twice or more|he was seen once)= P(he has hiked in this park twice or more AND he was seen once) / P(he was seen once)
= P(he has hiked in this park twice or more) / P(he was seen once)
Now, P(he has hiked in this park twice or more)
= 1 - P(he has hiked exactly once) - P(he does not hiked this park)
= 1 - C(10,1) * (1/6)^1 * (5/6)^9 - (5/6)^10
= 0.5155
P(he was seen once)
= 1 - P(he does not hiked this park)
= 1 - (5/6)^10
= 0.8385
So,
P(he has hiked in this park twice or more|he was seen once)
= 0.5155 / 0.8385
= 0.615
2007-01-31 2:43 am
Let T be the number of times he hiked in Oregon Ridge.
Required (conditional) probability
= P( T>1 l T>0 )
= P( T>1 and T>0 ) / P( T>0 )
= P( T>1 ) / P( T>0 )
= [ 1 - (5/6)^10 - 10C1(1/6)(5/6)^9 ] / [ 1 - (5/6)^10 ]
= 0.614772...
= 0.615 (cor to 3 d.p.)
Required (conditional) probability
= P( T>1 l T>0 )
= P( T>1 and T>0 ) / P( T>0 )
= P( T>1 ) / P( T>0 )
= [ 1 - (5/6)^10 - 10C1(1/6)(5/6)^9 ] / [ 1 - (5/6)^10 ]
= 0.614772...
= 0.615 (cor to 3 d.p.)
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070130000051KK02625