A family has j children with probability pj , where p1 = .1, p2 = .25, p3 = .35, p4 = .3. A child from this family is randomly chosen. Given that the child is the eldest child in the family, find the conditional probability that the family has

A) the conditional probability that the has only 1 child = 0.24

B) The conditional probability that the family has only 4 children = 0.18

Step-by-step explanation:

To answer the questions, we first start with defining each event. Let E be the event that the child selected is the oldest and let Fj be the event that the family has j children.

From this, we can deduce that the probability that the child is the oldest, given that there is j children is ; P(E | Fj ) = 1/j.

In addition, we know P(Fj ) = pj as given in the problem. In answering the 2 questions, we seek the probability P(Fj | E). Thus, by the Bayes’s formula;

P(Fj | E) = P(EFj )/P(E) which gives;

P(Fj | E) = {P(E | Fj )P(Fj )} / {Σ(4,i=1)P(E | Fj )P(Fj )}

= (1/j)/ Σ(4,i=1)(1/j)pj

= ((pj)/j) / {p1 + (p2)/2 + (p3)/3 +(p4)/4}

Therefore, the conditional probability that the family has only 1 child; P(F1 | E) = p1 / {p1 + (p2)/2 + (p3)/3 +(p4)/4} = 0.1/ (0.1 + (0.25/2) +(0.35/3) + (0.3/4) = 0.1/0.4167 = 0.24

The conditional probability that the family has only 4 children =

{(p4)/4} / {p1 + (p2)/2 + (p3)/3 +(p4)/4} = (0.3/4)/ (0.1 + (0.25/2) +(0.35/3) + (0.3/4) = 0.075/0.4167 = 0.18

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)

$15 each

step-by-step explanation:

$45/3 children= $15 each