can anyone show me how to work this problem out thanks in advance

The Census Bureau provides data on the number of young adults, ages 18–24, who are living in their parents' home (the data include single young adults who are living in college dormitories because it is assumed these young adults will return to their parents' home when school is not in session).

Let the variables M and F represent the following events.

1. M = the event a male young adult is living in his parents' home
F = the event a female young adult is living in her parents' home

If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude P(M) = .56 and P(F) = .42 (The World Almanac, 2006). The probability that both are living in their parents' home is .24.

What is the probability both young adults selected are living on their own (neither is living in their parents' home) (to 2 decimals)?

rfwd

To find the probability that both young adults selected are living on their own, we can use the concept of conditional probability. Conditional probability is the probability of an event occurring given that another event has already occurred.

Let's break down the problem step by step:

1. We are given that P(M) = 0.56, which represents the probability that a randomly selected male young adult is living in his parents' home. Similarly, we're given that P(F) = 0.42, which represents the probability that a randomly selected female young adult is living in her parents' home.

2. We're also given that the probability that both young adults are living in their parents' home (P(M ∩ F)) is 0.24. This means the probability of both events M and F happening together is 0.24.

Now, let's find the probability that both young adults selected are living on their own:

1. If we assume that young adults living in their parents' homes or independently are the only two possibilities, then the probability of living on their own can be expressed as 1 minus the probability of living with parents.

2. The probability that a male young adult is NOT living in his parents' home is 1 - P(M). Similarly, the probability that a female young adult is NOT living in her parents' home is 1 - P(F).

3. The probability that both young adults selected are living on their own can be calculated using the product of the probabilities of each individual not living in their parents' home. Thus, P(not M and not F) = (1 - P(M)) * (1 - P(F)).

Finally, substituting the given values into the formula, we have:

P(not M and not F) = (1 - 0.56) * (1 - 0.42) = 0.44 * 0.58 = 0.2552.

So, the probability that both young adults selected are living on their own is approximately 0.26 (rounded to 2 decimal places).