Final answer:
The probability that Jones has two sons given that he is invited to the father-son dinner is 1/4 or 0.25.
Step-by-step explanation:
The question asks for the conditional probability that Jones has two sons given that he is invited to the father-son dinner. To calculate this probability, we need to use the concept of conditional probability. Let's assume that the probability of having a boy or a girl is 1/2 each for each child. Since we know that Jones has two children and one of them is a son (since he is invited to the dinner), we can create a sample space of possible outcomes. There are four possible outcomes: BB, BG, GB, and GG, where the letters represent the gender of his two children. Out of these four outcomes, only one outcome (BB) satisfies the condition that both children are boys. Therefore, the probability that Jones has two sons given that he is invited to the dinner is 1/4 or 0.25.