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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 -

    Is this right for A)

    n=6
    p=P(.38)
    q= 1-.38=.62 q=P(.62)
    P(x=0)= 6C0= 6!/ 0!(6-0)!= 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?

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