Two integers are selected at random from the integers 1 to 11. If their sumis even find the proabability that both integers are odd.

Answer

N/a

Explanation

Here `n(S)=.^(11)C_(2)=55`
Let `E=` event of selecting two integers whose sum is even.
The sum of two integers will be even if either both integers are even or both integers are odd.
`:.n(E)=.^(6)C_(2)+.^(5)C_(2)=15+10=25`
Let `F=` event that both integers are odd
`impliesn(F)=.^(6)C_(2)=15`
`:.n(FnnE)=.^(6)C_(2)=15`
Now `P(F//E)=(P(FnnE))/(P(E))=(15//55)/(25//55)=3/5`