Posted by Sunny on .
Studies show that 38% of all women suffer from hip fractures caused by osteoporosis by age 85. If 6 women aged 85 are randomly selected, what is the probability that
A) None will suffer or has suffered a hip fracture P(x = 0)
B) At least 4 have suffered a hip fracture P(x ¡Ü 4)
C) Fewer than 2 have suffered a hip fracture P(x < 2)

Stats 
Sunny,
Is this right for A)
n=6
p=P(.38)
q= 1.38=.62 q=P(.62)
P(x=0)= 6C0= 6!/ 0!(60)!= 6!/0!(6)!=
6*5*4*3*2*1/1*6*5*4*3*2*1= 720/720= 1
6C0 (.38)0 (.62)6= (1)(1)(.0568)= .0568
so P(x=0)= .0568?