(New page: For problem 2 I think a) [i] P(one lost on A,one lost on B)=2!(1/5)(1-1/5)(1/16) [ii] P(both lost on B)=[(1-1/5)^2]*(1/16) [iii] P(both lost on A)=(1/5)^2 I don't really understand...) |
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For problem 2 | For problem 2 | ||
I think | I think | ||
| − | a) | + | a) |
| − | + | ||
| − | + | (i) P(one lost on A,one lost on B)=2!(1/5)(1-1/5)(1/16) | |
| + | |||
| + | (ii) P(both lost on B)=[(1-1/5)^2]*(1/16) | ||
| + | |||
| + | (iii) P(both lost on A)=(1/5)^2 | ||
I don't really understand the solutiong they give, but i dont really think the solution for a)[ii] is correct. The P for one case lost in B is 1/16, how can lost both in B even higher than lost one? | I don't really understand the solutiong they give, but i dont really think the solution for a)[ii] is correct. The P for one case lost in B is 1/16, how can lost both in B even higher than lost one? | ||
Latest revision as of 15:51, 24 September 2008
For problem 2 I think a)
(i) P(one lost on A,one lost on B)=2!(1/5)(1-1/5)(1/16)
(ii) P(both lost on B)=[(1-1/5)^2]*(1/16)
(iii) P(both lost on A)=(1/5)^2
I don't really understand the solutiong they give, but i dont really think the solution for a)[ii] is correct. The P for one case lost in B is 1/16, how can lost both in B even higher than lost one?
