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

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

Grade:12

2 Answers

abhay prakash
21 Points
7 years ago
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+11
n(S)=11+11+11..............11times=121
E={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+11
n(E)=6+6+6..........11times =66
P(E)=66/121
Ashish bedi
54 Points
7 years ago
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

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