## Elementary Statistics (12th Edition)

a) $P(X=13)={14\choose 13} \cdot (0.5)^{13}\cdot (0.5)^1=0.000854.$ b)$P(X=14)={14\choose 14} \cdot (0.5)^{14}\cdot (0.5)^0=0.000061.$ c)$P(X\geq13)=P(X=13)+P(X=14)={14\choose 13} \cdot (0.5)^{13}\cdot (0.5)^1+{14\choose 14} \cdot (0.5)^{14}\cdot (0.5)^0=0.000854+0.000061=0.000915.$ d)The probability of c) indicates that normally this probability would be very low therefore the method seems to be effective.