(New page: Category:ECE302Spring2013Boutin Category:ECE Category:ECE302 Category:probability problem solving Category:independence Q: There are five volunteers(Cal...) |
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| − | [[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:probability]] [[Category | + | [[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:probability]] [[Category:problem solving]] |
[[Category:independence]] | [[Category:independence]] | ||
| + | ==Practive Problem on Independence== | ||
Q: There are five volunteers(Called A,B,C,D,E), and there are four volunteer positions(Position 1, postion 2, position 3, position 4). A volunteer is equally likely to be place on one of these for positions. | Q: There are five volunteers(Called A,B,C,D,E), and there are four volunteer positions(Position 1, postion 2, position 3, position 4). A volunteer is equally likely to be place on one of these for positions. | ||
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= p(A is in position 1) *p(C and D is not in position 1)(* p(B is in position 1) | = p(A is in position 1) *p(C and D is not in position 1)(* p(B is in position 1) | ||
= (1/4)*(1/4)*(1-(1/16)) = (1/16)*(15/16) = 15/256 | = (1/4)*(1/4)*(1-(1/16)) = (1/16)*(15/16) = 15/256 | ||
| − | + | ---- | |
| − | + | ==Comments/questions== | |
| + | *Write question/comment here. | ||
| + | **answer here. | ||
| + | What he did is correct except a small calculation error => (1/4)^4 = 1/256, not 1/64. | ||
[[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | [[Bonus_point_1_ECE302_Spring2012_Boutin|Back to first bonus point opportunity, ECE302 Spring 2013]] | ||
Latest revision as of 09:57, 4 February 2013
Practive Problem on Independence
Q: There are five volunteers(Called A,B,C,D,E), and there are four volunteer positions(Position 1, postion 2, position 3, position 4). A volunteer is equally likely to be place on one of these for positions.
1.What is the probability that five volunteers are in the same position? 2.Given that C and D is not in position 1, what is the probability that both A and B are in the position 1?
A:
1.
P1 = p(all in position 1) + p(all in position 2) + p(all in position 3) + p(all in position 4)
= [(1/4)^4]*(1/4) + [(1/4)^4]*(1/4) + [(1/4)^4]*(1/4) + [(1/4)^4]*(1/4)
= (1/4)^4
= 1/64
2.
p2 = p(A is in position 1|C and D is not in position 1)*p(B is in position 2|C and D is not in position 1) Event A is in position 1 is independent with the Event C and D is in position 1.
= p(A is in position 1) *p(C and D is not in position 1)(* p(B is in position 1)
= (1/4)*(1/4)*(1-(1/16)) = (1/16)*(15/16) = 15/256
Comments/questions
- Write question/comment here.
- answer here.
What he did is correct except a small calculation error => (1/4)^4 = 1/256, not 1/64.
