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  • https://stats.libretexts.org/Sandboxes/JolieGreen/Finite_Mathematics_-_Spring_2023_-_OER/11%3A_Probability/11.02%3A_Mutually_Exclusive_Events_and_the_Addition_Rule
    Given two events, E, F, then finding the probability of E \(\cup\) F, is the same as finding the probability that E will happen, or F will happen, or both will happen. Let E be the event that the numb...Given two events, E, F, then finding the probability of E \(\cup\) F, is the same as finding the probability that E will happen, or F will happen, or both will happen. Let E be the event that the number shown on the die is an even number, and let F be the event that the number shown is greater than four. If we count the number of elements n(E) in E, and add to it the number of elements n(F) in F, the points in both E and F are counted twice, once as elements of E and once as elements of F.
  • https://stats.libretexts.org/Under_Construction/Purgatory/FCC_-_Finite_Mathematics_-_Spring_2023/11%3A_Probability/11.02%3A_Mutually_Exclusive_Events_and_the_Addition_Rule
    Given two events, E, F, then finding the probability of E \(\cup\) F, is the same as finding the probability that E will happen, or F will happen, or both will happen. Let E be the event that the numb...Given two events, E, F, then finding the probability of E \(\cup\) F, is the same as finding the probability that E will happen, or F will happen, or both will happen. Let E be the event that the number shown on the die is an even number, and let F be the event that the number shown is greater than four. If we count the number of elements n(E) in E, and add to it the number of elements n(F) in F, the points in both E and F are counted twice, once as elements of E and once as elements of F.
  • https://stats.libretexts.org/Courses/Fresno_City_College/New_FCC_DS_21_Finite_Mathematics_-_Spring_2023/11%3A_Probability/11.02%3A_Mutually_Exclusive_Events_and_the_Addition_Rule

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