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`        two integers are selected from integers 1to11. if sum is even, find the probability that both the numbers are odd`
2 years ago

## Answers : (2)

abhay prakash
21 Points
```							S={1+1,1+2,1+3,..............1+11,2+1,2+2,2+3.......,2+11,3+1,3+2.........,3+11,4+1,4+2.....,4+11,5+1,......5+11,6+1,....6+11,7+1,.....7+11,8+1,.......8+11,9+1,...9+11,10+1,.....10+11,11+1,....11+11n(S)=11+11+11..............11times=121E={1+1,1+3,1+5,1+7,1+9,1+11,3+1,3+3,3+5,3+7,3+9,3+11,5+1,5+3,5+5,5+7,5+9,5+11,............................11+11n(E)=6+6+6..........11times =66P(E)=66/121
```
2 years ago
Ashish bedi
54 Points
```							Answer is 3/5. Apply baes theorem. We can  have 2 odd or 2evens for there sum even.  So total probability will be 6c2/11c2 divided by 6c2/11c2+5c2/11c2. =3/5
```
2 years ago
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