Question

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is oyer 6 ft in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A) ?\) Why?

Short answer

The probability of an adult male being over 6 ft in height given that he is a professional basketball player (\(P(A|B)\)) is more likely to be larger than the probability of an adult male being a professional basketball player given that he is over 6 ft (\(P(B|A)\)).

Step-by-step solution

Understanding the Concept of Conditional Probability

Conditional probability is the probability of an event given that another event has occurred. In this case, \(P(A|B)\) denotes the probability of 'A' happening given that 'B' has already happened. Similarly, \(P(B|A)\) denotes the probability of 'B' happening given that 'A' has already happened.

Relate to Real-world Scenario

In the context of the problem, \(P(A|B)\) refers to the probability of an adult male being over 6 ft in height given that he is already confirmed to be a professional basketball player. This is likely to be high because the majority of professional basketball players are tall. On the other hand, \(P(B|A)\) refers to the probability that an adult male is a professional basketball player given that he is known to be over 6ft. While many professional basketball players are indeed tall, not all tall individuals are professional basketball players.

Conclusion

Therefore, it is highly probable to suggest that \(P(A|B)\) is higher than \(P(B|A)\) because the probability of a professional basketball player being over 6ft is higher than the probability of a tall person being a professional basketball player.