The probability of having a one-child family is $0.1$; the probability of having a two child family is $0.25$; the probability of having a three- child family is $0.35$ and the probability of having a four-child family is $0.3$. If a randomly chosen child is the eldest child in his/her family, calculate the probability that he/she is from:
(a) a one-child family;
(b) a four-child family
my thoughts for part a) 0.1/10 = 0.01 or 1% and for part b) 0.3/10=0.03 or 3%. Unless I am missing something here…
Best Answer
Use Bayes' rule, where $e$ represents "eldest is chosen":
$$P(1|e) = \frac{P(e|1)P(1)}{P(e)} = \frac{P(e|1)P(1)}{\sum\limits_{i=1}^4 P(e|i)P(i)} \\ = \frac{1 \cdot 0.1}{0.1 \cdot 1 + 0.25 \cdot 1/2 + 0.35 \cdot 1/3 + 0.3 \cdot 1/4} \\ = \frac{0.1}{0.416667} = 0.24 .$$