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)
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?
To find the probabilities, we need to use the binomial probability formula, which is expressed as:
P(x) = (n C x) * p^x * q^(n-x)
Where:
- P(x) is the probability of getting x successes
- (n C x) is the number of combinations of choosing x items from a set of n items, calculated as n! / (x! * (n-x)!)
- p is the probability of success for an individual trial
- q is the probability of failure for an individual trial, calculated as 1 - p
- n is the total number of trials
Given that 38% of women suffer from hip fractures by age 85, the probability of an individual woman suffering a hip fracture is p = 0.38. Therefore, the probability of an individual woman not suffering a hip fracture is q = 1 - 0.38 = 0.62.
Now let's calculate the probabilities:
A) None of the 6 women will suffer or have suffered a hip fracture (P(x = 0)):
In this case, x = 0, n = 6, p = 0.38, and q = 0.62.
Using the formula, we have:
P(x = 0) = (6 C 0) * (0.38^0) * (0.62^6)
= 1 * 1 * 0.062
= 0.062
So, the probability that none of the 6 women will suffer or have suffered a hip fracture is 0.062 or 6.2%.
B) At least 4 of the 6 women have suffered a hip fracture (P(x ≤ 4)):
In this case, we need to calculate the probabilities for x = 4, 5, and 6 and sum them.
Using the formula:
P(x = 4) = (6 C 4) * (0.38^4) * (0.62^2)
= 15 * 0.03856 * 0.3844
= 0.219615
P(x = 5) = (6 C 5) * (0.38^5) * (0.62^1)
= 6 * 0.0146416 * 0.62
= 0.05482976
P(x = 6) = (6 C 6) * (0.38^6) * (0.62^0)
= 1 * 0.00775984 * 1
= 0.00775984
Now, summing these probabilities:
P(x ≤ 4) = P(x = 4) + P(x = 5) + P(x = 6)
= 0.219615 + 0.05482976 + 0.00775984
= 0.2822046
So, the probability that at least 4 of the 6 women have suffered a hip fracture is 0.2822046 or 28.2%.
C) Fewer than 2 of the 6 women have suffered a hip fracture (P(x < 2)):
In this case, we need to calculate the probabilities for x = 0 and x = 1 and sum them.
Using the formula:
P(x = 0) = 0.062 (calculated in part A)
P(x = 1) = (6 C 1) * (0.38^1) * (0.62^5)
= 6 * 0.38 * 0.077414
= 0.17760752
Now, summing these probabilities:
P(x < 2) = P(x = 0) + P(x = 1)
= 0.062 + 0.17760752
= 0.23960752
So, the probability that fewer than 2 of the 6 women have suffered a hip fracture is 0.23960752 or 23.9%.