The probability is 0.337.
The probability that more than 3 prefer brand c is the same as finding the probability that 4 or 5 prefer brand c. Using a binomial distribution,
[tex]_5C_4(0.6)^4(0.4)^{5-4}+_5C_5(0.6)^5(0.4)^{5-5}
\\
\\=\frac{5!}{4!1!}(0.6)^4(0.4)^1+\frac{5!}{5!0!}(0.6)^5(0.4)^0
\\
\\=5(0.6)^4(0.4)+1(0.6)^5(1)=0.33696\approx0.337[/tex]