Afamily has two children. if b represents a boy and g represents a girl, the set of outcomes for the possible genders of the children is s = {bb, bg, gb, gg}, with the oldest child listed first in each pair. let x represent the number of times g occurs. which of the following is the probability distribution, px(x)?

P(X=0) = 1/4,  P(X=1)=1/2 and  P(X=2)= 1/4.

Step-by-step explanation:

Since, the possible genders of the children is S = {BB, BG, GB, GG}

Here, Let X represent the number of times G occurs.

Thus, when there is no girl

Possible arrangement = {BB}

Then P(X=0) = 1/4

When there is one girl,

Then possible arrangement = { BG, GB}

Thus P(X=1)= 2/4

⇒ P(X=1)=1/2

When there  is 2 girls,

Then possible arrangement = {GG}

Thus P(X=2)= 1/4

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

a) justify

step-by-step explanation:

justifying is just part of everyday life. justify kind of means proof so i think the answer is a) justify.