But let us assume that it is true.
The probability that a random reader has read that ad. is 1-(1-8/10)(1-7/10)(1-6/10) ~=0.976.
Now we can treat this as constant for all readers (because the number of readers is large and the number of non-ad. readers realtivly small)
Therefore we may treat the number of readers as a binomial random variable with mean 9.76 million and standard deviation ~=484. Now we can use the normal approximation to evaluate the required probability, which will be negligable.
There may be a better way of doing the calculation buut it will give the same sort of result.