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Is it true the special rule of addition is the events must be mutually exclusive?
yes
What important question must you answer before computing an or probability?
You need to know whether or not the events are mutually exclusive.
Which one is more convienent mutually exclusive or non mutually?
It must be "mutually exclusive" since "non mutually" does not even mean anything!
Can a probability distribution be a mutually exclusive listing of the outcomes of an experiment which can occur by chance and the corresponding probabilities of occurrance?
Not quite. The listing must also be exhaustive: it must contain all possible outcomes.For the roll of a fair cubic die, consider the following:Prob(1) = 1/6Prob(2) = 1/6This is a mutually exclusive listing of the outcomes of the experiment and the corresponding probabilities of occurrence but it is not a probability distribution because it does not include all possible outcomes. As a result, the total of the listed probabilities is less than 1.
A six sided die is rolled twice what is the probability of rolling either an even number on the first roll or a 1 on the second roll the answer is 7 over 12 but I can't figure out how to get there?
The question asks "What is the probability of rolling either an even number on the first roll or a 1 on the second roll?" These events are independent from each other as the outcome of the second roll is not affected by the outcome of the first roll. However, these events are non-mutually exclusive, meaning that these events can both occur at the same time.The probability of rolling an even number on the first roll is 3/6 because 2, 4, and 6 are even numbers and a six-sided die has six possible numbers.The probability of rolling a 1 on the second roll is 1/6.If these two probabilities are added together, we will have "double counted" the event where an even number is rolled on the first roll and a 1 is rolled on the second roll. To correct for this, we must subtract the probability of both events occurring.The probability that both events occur is 3/36, because 3/6 * 1/6 = 3/36.Now, the probability of rolling either an even number on the first roll or a 1 on the second roll is:3/6 + 1/6 - 3/36= 18/36 + 6/36 - 3/36= 21/36= 7/12
