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
