The simple-minded approach would be to calculate the expected number of non-smokers: (n)(p) = (120)(0.6) = 72.

But how sure is 'very sure'? 99% sure? Having 72 seats wouldn't make you 99% sure ...... (in fact, it would only make you about 53% sure!).

Let X be the random variablenumber of non-smoking seats required.

Then X ~ Binomial (n = 120, p = 0.6)

To be say, 99% sure, you need to find the smallest value of x such that . This value of x is the number of seats required.

You can use technology to very easily solve for x ..... I get x = 84.