Lesson Dependent and independent events REVISITED
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<H2>Dependent and independent events REVISITED</H2> <H3>Problem 1</H3>In a pet store, a customer is randomly selected. It was determined that the probability that they owned a dog was 60% and the probability they owned a cat was 50%. The probability that the customer owner both a cat and a dog was 23%. Are the events “owning a cat” and “owning a dog” independent or dependent events? <B>Solution</B> <pre> According to the definition of <U>independent events</U>, we should check if P(owning a dog AND owning the cat) = P(owning a dog)*P(owning a cat). The left side is 0.23 (given). The right side is 0.6*0.5 = 0.3. The two sides values are different, so the events ARE NOT INDEPENDENT. <U>ANSWER</U> </pre> <H3>Problem 2</H3>The probability of snow falling on a typical day of December is 0.10 and the probability of a traffic jam is 0.80. The probability of snow or a traffic jam is 0.82. Are the events "it snows" and "a traffic jam occurs" independent ? <B>Solution</B> <pre> From the basic formula of the elementary probability formula P(snow OR a traffic jam) = P(snow) + P(a traffic jam) - P(snow AND a traffic jam). (1) Substitute the given data into this formula. You will get 0.82 = 0.8 + 0.10 - P(snow AND a traffic jam). (2) From this equation P(snow AND a traffic jam) = 0.8 + 0.1 - 0.82 = 0.08. (3) From the other hand side, P(snow) * P(a traffic jam) = 0.1 * 0.8 = 0.08. (4) From (3) and (4), we see that P(snow AND a traffic jam) = P(snow) * P(a traffic jam). 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