Answer
$P(fewer~than~5)=0.00935$
Work Step by Step
$P(fewer~than~5)=P(X\lt5)=P(0)+P(1)+P(2)+P(3)+P(4)={}_{15}C_{0}~0.6^0~(0.4)^{15}+{}_{15}C_{1}~0.6^1~(0.4)^{14}+{}_{15}C_{2}~0.6^2~(0.5)^{13}+{}_{15}C_{3}~0.6^3~(0.4)^{12}+{}_{15}C_{4}~0.6^4~(0.5)^{11}=\frac{15!}{0!\times(15-0)!}\times1\times0.4^{15}+\frac{15!}{1!\times(15-1)!}\times0.6\times0.4^{14}+\frac{15!}{2!\times(15-2)!}\times0.36\times0.4^{13}+\frac{15!}{3!\times(15-3)!}\times0.216\times0.4^{12}+\frac{15!}{4!\times(15-4)!}\times0.1296\times0.4^{11}=1\times1\times0.4^{15}+15\times0.6\times0.4^{14}+105\times0.36+455\times0.216\times0.4^{12}+1365\times0.1296\times0.4^{11}=0.00935$
