Final answer:
In this case, the probability of having dinner with his mother is indeed 0.5, as Marvin claims. However, the probability of having dinner with her twice in 20 working days is only around 5.37%. Therefore, it is statistically unlikely for Marvin to have dinner with his mother frequently based on random subway choices.
Step-by-step explanation:
The problem of "The U`nfair Subway" involves Marvin, who gets off work at random times between 3 and 5 p.m. His mother lives uptown, his g`irlfriend downtown. He takes the first subway that comes in either direction and eats dinner with the one he is delivered to. His mother complains that he never comes to see her, but he says she has a 50-50 chance. He has had dinner with her twice in the last 20 working days.
In this scenario, Marvin's claim of a 50-50 chance of having dinner with his mother is not accurate. To understand why, let's analyze the situation using probability.
1. Determine the probability of Marvin going uptown: Since Marvin's mother lives uptown, he can have dinner with her only when he takes the subway in that direction. As there is an equal chance of the subway going uptown or downtown, the probability of Marvin going uptown is 1/2 or 0.5.
2. Calculate the probability of not going uptown: The probability of Marvin not going uptown is the complement of going uptown, which is 1 - 0.5 = 0.5.
3. Calculate the probability of having dinner with his mother: Marvin can have dinner with his mother only when he goes uptown. Since the probability of going uptown is 0.5, the probability of having dinner with his mother is also 0.5.
4. Determine the probability of having dinner with his mother twice in 20 working days: This probability can be calculated using the binomial probability formula.
Let's assume the probability of having dinner with his mother on any given working day is p = 0.5.
The probability of having dinner with his mother twice in 20 working days can be calculated as:
P(X = 2) = (20 choose 2) * (0.5)² * (0.5)⁽²⁰⁻²⁾
where (20 choose 2) represents the number of ways to choose 2 out of 20 working days.
P(X = 2) = 190 * (0.5)²* (0.5)¹⁸ ≈ 0.0537
Therefore, the probability of having dinner with his mother twice in 20 working days is approximately 0.0537 or 5.37%.
We can see from the computation above that the likelihood of eating supper with his mother is 0.5, as Marvin asserts. However, eating supper with her twice in 20 working days is just about 5.37% likely. As a result, based on random subway choices, Marvin is statistically unlikely to have dinner with his mother on a regular basis.
Your question is incomplete, but most probably the full question was:
Can you solve the problem of "The Unfair Subway"?
Marvin gets off work at random times between 3 and 5 p.m. His mother lives uptown, his gi`rlfriend downtown. He takes the first subway that comes in either direction and eats dinner with the one he is delivered to. His mother complains that he never comes to see her, but he says she has a 50-50 chance. He has had dinner with her twice in the last 20 working days. Explain.
Marvin's adventures in probability are one of the fifty intriguing puzzles that illustrate both elementary ad advanced aspects of probability, each problem designed to challenge the mathematically inclined. From "The Flippant Juror" and "The Prisoner's Dilemma" to "The Cliffhanger" and "The Clumsy Chemist," they provide an ideal supplement for all who enjoy the stimulating fun of mathematics. Professor Frederick Mosteller, who teaches statistics at Harvard University, has chosen the problems for originality, general interest, or because they demonstrate valuable techniques.
In addition, the problems are graded as to difficulty and many have considerable stature. Indeed, one has "enlivened the research lives of many excellent mathematicians."
Detailed solutions are included. There is every probability you'll need at least a few of them.