Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the probability of a household having: (a) 2 or 5 children (b) 3 or fewer children (c) 8 or more children (d) fewer than 5 children (e) more than 3 children

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

In this problem, we have that:

(a) 2 or 5 children

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

"if it is my birthday, then it is snowing"

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option