Let A=%ge reading morning paper only, B=%ge reading evening newspaper only, C=%ge reading both. C=20%.
Those reading the morning papers include those reading both: A+C=50%, so A+20=50%, and A=30%. Those reading the evening papers include those reading both: B+C=60, B+20=60, B=40%.
Let N be those residents reading neither newspapers, then A+B+C+N=100%, so N=100-(30+40+20)=10%.
Since 10% of residents read neither morning nor evening newspapers the remaining 90% must read either or both types of newspapers. So the probability is 90%.