Coughlin et al. estimated the percentage of women living in border countries along the southern united states with mexico who have less than high school education to be 18.7. Assume the corresponding probability is .19. Suppose we select three women at random. Find the probability that the number with less than a high school education is:

a. exactly zero

b. exactly one

c. more than one

d. two or fewer

e. two or three

f. exactly three

Relevance

Hi,

Coughlin et al. estimated the percentage of women living in border countries along the southern united states with mexico who have less than high school education to be 18.7. Assume the corresponding probability is .19. Suppose we select three women at random. Find the probability that the number with less than a high school education is:

a. exactly zero

3 nCr 0(.19)^0 (.81)^3 = .53144 <==ANSWER

b. exactly one

3 nCr 1(.19)^1 (.81)^2 = .373977 <==ANSWER

c. more than one

3 nCr 2(.19)^2 (.81)^1 + 3 nCr 3(.19)^3 (.81)^0 = .087723 + .006859 = .094582 <==ANSWER

d. two or fewer

3 nCr 0(.19)^0 (.81)^3 + 3 nCr 1(.19)^1 (.81)^2 + 3 nCr 2(.19)^2 (.81)^1 = .53144 + .373977 + .087723 = .99314 <==ANSWER

e. two or three

3 nCr 2(.19)^2 (.81)^1 + 3 nCr 3(.19)^3 (.81)^0 = .087723 + .006859 = .094582 <==ANSWER

f. exactly three

3 nCr 3(.19)^3 (.81)^0 = .006859 <==ANSWER

I hope that helps!! :-)

• Anonymous
9 years ago

I think you do it like this, but not sure! Probability of the thing you're looking for is 0.19, probability of the other is therefore 0.81.

a) 0.81 x 0.81 x 0.81 = 0.531

b) 0.19 x 0.81 x 0.81 = 0.125

g) (exactly two) 0.19 x 0.19 x 0.81 = 0.029 (not asked but makes the rest easier)

f) 0.19 x 0.19 x 0.19 = 0.007

c) = g) + f) = 0.036

d) = a) + b) + g) = 0.685

e) = g) + f) = 0.036

I have assumed that there are so many women that taking the three out doesn't make a significant difference to the total and therefore teh probabilities.

I would actually love to know if this is correct!