(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...) |
|||
| Line 1: | Line 1: | ||
For problem 2 | For problem 2 | ||
I think | I think | ||
| − | a) [i] P(one lost on A,one lost on B)=2!(1/5)(1-1/5)(1/16) | + | 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? | ||
Revision as of 15:50, 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?
