Question

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

Answer 1

Answer:

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