In many families both parents work, as a result there is increasing need for day care. Data was collected; and one year in Canada approximately 32% of children aged 0 to 11 years were in day care for at least 20h per week.

Question

In many families both parents work, as a result there is increasing need for day care. Data was collected; and one year in Canada approximately 32% of children aged 0 to 11 years were in day care for at least 20h per week. 
What is the probability, in a random poll of 60 children from the ages of 0 to 11 that more than 15 children are in day care at least 20h per week? 
p(k)= nCk P^k (1-p)^n-k 
were 
n= 60 
p=0.32 
k= 15 
q= 1-0.32= 0.68 
so: 
p(k) = 60C15 x 0.32^15 (0.68) ^60-15 
= 0.0583464079 
is this correct?

What is the probability in a random poll of 60 children that fewer than 20 are in a day care at least 20h per week? 
How would I do this because it says fewer?

Answer is incorrect.

You could answer this exactly by using a combinatorial application as you have tried to do, or you could approximate the answer using the properties of a binominal distribution. I will use the latter. (The formula you gave is the would give the probability of observing EXACTLY 15 children.)

With 60 kids we would expect .32*60=19.2 kids to be in day care. This estimate has an estimated standard deviation of sqrt(n*p*q) = sqrt(60*.32*.68) = 3.613. Now then you want to know the likelyhood of observing more than 15 kids. (I will use a mid-point of 15.5) So, 19.2-15.5 = 3.7. This is (3.7/3.613)=1.02 standard deviations below the mean. Looking in my Cumulative Normal Distribution table, 1.02 corresponds to .8461. ERGO, there is approximately an 85% chance of observing more than 15 children out of a sample of 60 will be in day care.

Follow the same steps for estimating the number fewer than 20. It should be slightly greater than 50%.