You can't. To answer this you are probably expected to assume that the ad. readers on any day are an independent sample of all 10 million readers, which is very unlikely to be true.

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.

CB