Sociologists say that 90% of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). Suppose that six married women are having coffee together one morning. Find the following probabilities. (For each answer, enter a number. Round your answers to three decimal places.)

The number of women out of 12 selected dislike their mother-in-law is a random variable that can be represented by X. Then X is a Binomial distribution with

P( dislike their mother-in-law) p = 0.90

n= 12

Therefore, the probability distribution of X is P(x)= nCxp^x(1-p^(n-x) x=0,1,. ... ,12

1) P(all of them dislike their mother-in-law)= P(X=12) = (12C12)(0.90)^12(0.10)^0

= 0.2824

2) P(none of them dislike their mother-in-law) =P( X=0) = 12C0(0.90)^0(0.10)^12

=1x10^(-12)

3) P(at least ten of them dislike their mother-in-law)= P(X>=1)=1-P(X=0)

=1-1x10^(-12)=0.9999999999999

4)P(no more than two of them dislike their mother-in-law) =P( X<=2)

=p(X=0)+P(X=1)+P(X=2)= 5.45x10^(-9)

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}