Question

Two roommates share a dorm in Richards Hall, The probability that Roommate A is doing homework on a Friday night is .3. The probability that Roommate B is doing homework on a Friday night is .4. The probability that both roommates are doing homework on a Friday night is .2 (notice the two roommates are able to coordinate their decisions, so you can't assume their respective probabilities are independent). Find the probability that: At least one roommate is doing homework this Friday night

Answer 1

Answer:

There is a 50% probability that at least one roommate is doing homework this Friday night.

Step-by-step explanation:

This problem can be solved building the Venn Diagram of these probabilities.

I am going to say that P(A) is the probability that the roommate A is doing homework and P(B) is the probability that the roommate B is doing homework.

We have that:

$P(A) = P(a) + P(A \cap B)$

In which P(a) is the probability that only the roommate A is doing homework and $P(A \cap B)$ is the probability that both student A and student B are doing homework.

We also have that:

$P(B) = P(b) + P(A \cap B)$

The problem states that

The probability that Roommate A is doing homework on a Friday night is .3. So $P(A) = 0.3$.

The probability that Roommate B is doing homework on a Friday night is .4. So $P(B) = 0.4$

The probability that both roommates are doing homework on a Friday night is .2. So $P(A \cap B) = 0.2$

Find the probability that: At least one roommate is doing homework this Friday night

This is the probability that either only A is doing, either only B, or both. So:

$P = P(a) + P(b) + P(A \cap B)$

We have that

$P(A) = P(a) + P(A \cap B)$

We have P(A) and $P(A \cap B)$, so we can find P(a)

$P(A) = P(a) + P(A \cap B)$

$0.3 = P(a) + 0.2$

$P(a) = 0.1$

Also

$P(B) = P(b) + P(A \cap B)$

$0.4 = P(b) + 0.2$

$P(b) = 0.2$

So:

$P = P(a) + P(b) + P(A \cap B)$

$P = 0.1 + 0.2 + 0.2$

$P = 0.5$

There is a 50% probability that at least one roommate is doing homework this Friday night.