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The 12 unions that represent all of the more than 100,000 workers across the industry said Friday that collectively the six biggest freight railroads spent over $165 billion on buybacks well . But first, a definition: Probability of an event happening = Are \(\text{G}\) and \(\text{H}\) independent? These two events are independent, since the outcome of one coin flip does not affect the outcome of the other. The outcomes HT and TH are different. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in Part c is the number of outcomes (size of the sample space). Check whether \(P(\text{L|F})\) equals \(P(\text{L})\). The probabilities for \(\text{A}\) and for \(\text{B}\) are \(P(\text{A}) = \dfrac{3}{4}\) and \(P(\text{B}) = \dfrac{1}{4}\). Independent and mutually exclusive do not mean the same thing. You do not know \(P(\text{F|L})\) yet, so you cannot use the second condition. .3 then you must include on every digital page view the following attribution: Use the information below to generate a citation. To show two events are independent, you must show only one of the above conditions. I've tried messing around with each of these axioms to end up with the proof statement, but haven't been able to get to it. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about \(P(\text{Shirt} \#133|\leq 210 \text{ pounds})\)? 1 Are \(\text{C}\) and \(\text{E}\) mutually exclusive events? A box has two balls, one white and one red. Question: A) If two events A and B are __________, then P (A and B)=P (A)P (B). P(D) = 1 4 1 4; Let E = event of getting a head on the first roll. For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Since A has nothing to do with B (because they are independent events), they can happen at the same time, therefore they cannot be mutually exclusive. You put this card aside and pick the third card from the remaining 50 cards in the deck. A and B are mutually exclusive events if they cannot occur at the same time. Show \(P(\text{G AND H}) = P(\text{G})P(\text{H})\). Hearts and Kings together is only the King of Hearts: But that counts the King of Hearts twice! (This implies you can get either a head or tail on the second roll.) For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Start by listing all possible outcomes when the coin shows tails (.