Questions: A recent survey that 85% of a population watches TV at least once a day, 35% of the population uses a computer at least once a day, and 25% of the population do both. What is the probability that a person chosen at random from the population watches TV at least once a day or uses a computer at least once a day?

Solution Steps

Let $A$ be the event that a person watches TV at least once a day, and let $B$ be the event that a person uses a computer at least once a day. We are given the following probabilities:

• $P(A) = 0.85$
• $P(B) = 0.35$
• $P(A \cap B) = 0.25$

We need to find the probability that a person watches TV at least once a day or uses a computer at least once a day, which is the probability of the union of events $A$ and $B$, denoted as $P(A \cup B)$.

The formula for the probability of the union of two events is:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Substitute the given probabilities into the formula:

$$P(A \cup B) = 0.85 + 0.35 - 0.25$$

Calculate the probability:

$$P(A \cup B) = 0.85 + 0.35 - 0.25 = 0.95$$

Final Answer