{"id":46,"date":"2021-06-22T15:30:13","date_gmt":"2021-06-22T15:30:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/independent-and-mutually-exclusive-events\/"},"modified":"2023-12-05T09:01:18","modified_gmt":"2023-12-05T09:01:18","slug":"independent-and-mutually-exclusive-events","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/independent-and-mutually-exclusive-events\/","title":{"raw":"Probabilities of Two Events","rendered":"Probabilities of Two Events"},"content":{"raw":"
\r\n

Learning Outcomes<\/h3>\r\n
\r\n
    \r\n \t
  • Determine if sampling was done with or without replacement<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nIndependent and mutually exclusive do not<\/strong> mean the same thing.\r\n

    Independent Events<\/h3>\r\nTwo events are independent if the following are true:\r\n
      \r\n \t
    • [latex]P[\/latex]([latex]A[\/latex]|[latex]B[\/latex]) = [latex]P[\/latex]([latex]A[\/latex])<\/li>\r\n \t
    • [latex]P[\/latex]([latex]B[\/latex]|[latex]A[\/latex]) = [latex]P[\/latex]([latex]B[\/latex])<\/li>\r\n \t
    • [latex]P[\/latex]([latex]A[\/latex] AND [latex]B[\/latex]) = [latex]P[\/latex]([latex]A[\/latex])[latex]P[\/latex]([latex]B[\/latex])<\/li>\r\n<\/ul>\r\nTwo events\u00a0[latex]A[\/latex] and [latex]B[\/latex] are independent<\/strong> if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one<\/strong> of the above conditions. If two events are NOT independent, then we say that they are dependent<\/strong>.\r\n\r\nSampling may be done\u00a0with replacement<\/strong> or without replacement<\/strong>.\r\n
        \r\n \t
      • With replacement:<\/strong> If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.<\/li>\r\n \t
      • Without replacement:<\/strong> When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.<\/li>\r\n<\/ul>\r\nIf it is not known whether\u00a0[latex]A[\/latex] and [latex]B[\/latex] are independent or dependent, assume they are dependent until you can show otherwise<\/strong>.\r\n
        \r\n

        Example<\/h3>\r\nYou have a fair, well-shuffled deck of [latex]52[\/latex] cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are [latex]13[\/latex] cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit.\r\n
          \r\n \t
        1. Sa<\/strong>mpling with replacement:\u00a0<\/strong>Suppose you pick three cards with replacement. The first card you pick out of the [latex]52[\/latex] cards is the [latex]Q[\/latex] of spades. You put this card back, reshuffle the cards and pick a second card from the [latex]52[\/latex]-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the [latex]52[\/latex]-card deck. This time, the card is the [latex]Q[\/latex] of spades again. Your picks are {[latex]Q[\/latex] of spades, ten of clubs, [latex]Q[\/latex] of spades}. You have picked the [latex]Q[\/latex] of spades twice. You pick each card from the [latex]52[\/latex]-card deck.<\/li>\r\n \t
        2. Sampling<\/strong> without replacement:\u00a0<\/strong>Suppose you pick three cards without replacement. The first card you pick out of the [latex]52[\/latex] cards is the [latex]K[\/latex] of hearts. You put this card aside and pick the second card from the [latex]51[\/latex] cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining [latex]50[\/latex] cards in the deck. The third card is the [latex]J[\/latex] of spades. Your picks are {[latex]K[\/latex] of hearts, three of diamonds, [latex]J[\/latex] of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.<\/li>\r\n<\/ol>\r\n<\/div>\r\n

          <\/h3>\r\n
          \r\n

          Try it<\/h3>\r\nYou have a fair, well-shuffled deck of [latex]52[\/latex] cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are [latex]13[\/latex] cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit. Three cards are picked at random.\r\n
            \r\n \t
          1. Suppose you know that the picked cards are [latex]Q[\/latex] of spades, [latex]K[\/latex] of hearts and [latex]Q[\/latex] of spades. Can you decide if the sampling was with or without replacement?<\/li>\r\n \t
          2. Suppose you know that the picked cards are [latex]Q[\/latex] of spades, [latex]K[\/latex] of hearts, and [latex]J[\/latex] of spades. Can you decide if the sampling was with or without replacement?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124075\"]\r\nSolution:\r\n
              \r\n \t
            1. With replacement<\/li>\r\n \t
            2. No<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

              <\/h3>\r\n
              \r\n

              Example<\/h3>\r\nYou have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit. [latex]S[\/latex] = spades, [latex]H[\/latex] = hearts, [latex]D[\/latex] = diamonds, [latex]C[\/latex] = clubs.\r\n
                \r\n \t
              1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are [latex]QS[\/latex], [latex]1D[\/latex], [latex]1C[\/latex], [latex]QD[\/latex].<\/li>\r\n \t
              2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are [latex]KH[\/latex], [latex]7D[\/latex], [latex]6D[\/latex], [latex]KH[\/latex].<\/li>\r\n<\/ol>\r\nWhich of [latex]1[\/latex] or [latex]2[\/latex] did you sample with replacement and which did you sample without replacement?\r\n[reveal-answer q=\"124076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124076\"]\r\nSolution:\r\n
                  \r\n \t
                1. Without replacement<\/li>\r\n \t
                2. With replacement<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\n
                  \r\n\r\nThis video provides a brief lesson on finding the probability of independent events.\r\n\r\nhttps:\/\/youtu.be\/QsnfXUqsFPU\r\n

                  <\/h4>\r\n
                  \r\n

                  try it<\/h3>\r\nYou have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit. [latex]S[\/latex] = spades, [latex]H[\/latex] = hearts, [latex]D[\/latex] = diamonds, [latex]C[\/latex] = clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.\r\n
                    \r\n \t
                  1. [latex]QS[\/latex], [latex]1D[\/latex], [latex]1C[\/latex], [latex]QD[\/latex]<\/li>\r\n \t
                  2. [latex]KH[\/latex], [latex]7D[\/latex], [latex]6D[\/latex], [latex]KH[\/latex]<\/li>\r\n \t
                  3. [latex]QS[\/latex], [latex]7D[\/latex], [latex]6D[\/latex], [latex]KS[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124077\"]\r\n\r\nWithout replacement:\r\n
                      \r\n \t
                    1. Possible<\/li>\r\n \t
                    2. Impossible<\/li>\r\n \t
                    3. Possible<\/li>\r\n<\/ol>\r\nWith replacement:\r\n
                        \r\n \t
                      1. Possible<\/li>\r\n \t
                      2. Possible<\/li>\r\n \t
                      3. Possible<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"
                        \n

                        Learning Outcomes<\/h3>\n
                        \n
                          \n
                        • Determine if sampling was done with or without replacement<\/li>\n<\/ul>\n<\/section>\n<\/div>\n

                          Independent and mutually exclusive do not<\/strong> mean the same thing.<\/p>\n

                          Independent Events<\/h3>\n

                          Two events are independent if the following are true:<\/p>\n

                            \n
                          • [latex]P[\/latex]([latex]A[\/latex]|[latex]B[\/latex]) = [latex]P[\/latex]([latex]A[\/latex])<\/li>\n
                          • [latex]P[\/latex]([latex]B[\/latex]|[latex]A[\/latex]) = [latex]P[\/latex]([latex]B[\/latex])<\/li>\n
                          • [latex]P[\/latex]([latex]A[\/latex] AND [latex]B[\/latex]) = [latex]P[\/latex]([latex]A[\/latex])[latex]P[\/latex]([latex]B[\/latex])<\/li>\n<\/ul>\n

                            Two events\u00a0[latex]A[\/latex] and [latex]B[\/latex] are independent<\/strong> if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one<\/strong> of the above conditions. If two events are NOT independent, then we say that they are dependent<\/strong>.<\/p>\n

                            Sampling may be done\u00a0with replacement<\/strong> or without replacement<\/strong>.<\/p>\n

                              \n
                            • With replacement:<\/strong> If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.<\/li>\n
                            • Without replacement:<\/strong> When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.<\/li>\n<\/ul>\n

                              If it is not known whether\u00a0[latex]A[\/latex] and [latex]B[\/latex] are independent or dependent, assume they are dependent until you can show otherwise<\/strong>.<\/p>\n

                              \n

                              Example<\/h3>\n

                              You have a fair, well-shuffled deck of [latex]52[\/latex] cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are [latex]13[\/latex] cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit.<\/p>\n

                                \n
                              1. Sa<\/strong>mpling with replacement:\u00a0<\/strong>Suppose you pick three cards with replacement. The first card you pick out of the [latex]52[\/latex] cards is the [latex]Q[\/latex] of spades. You put this card back, reshuffle the cards and pick a second card from the [latex]52[\/latex]-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the [latex]52[\/latex]-card deck. This time, the card is the [latex]Q[\/latex] of spades again. Your picks are {[latex]Q[\/latex] of spades, ten of clubs, [latex]Q[\/latex] of spades}. You have picked the [latex]Q[\/latex] of spades twice. You pick each card from the [latex]52[\/latex]-card deck.<\/li>\n
                              2. Sampling<\/strong> without replacement:\u00a0<\/strong>Suppose you pick three cards without replacement. The first card you pick out of the [latex]52[\/latex] cards is the [latex]K[\/latex] of hearts. You put this card aside and pick the second card from the [latex]51[\/latex] cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining [latex]50[\/latex] cards in the deck. The third card is the [latex]J[\/latex] of spades. Your picks are {[latex]K[\/latex] of hearts, three of diamonds, [latex]J[\/latex] of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.<\/li>\n<\/ol>\n<\/div>\n

                                <\/h3>\n
                                \n

                                Try it<\/h3>\n

                                You have a fair, well-shuffled deck of [latex]52[\/latex] cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are [latex]13[\/latex] cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit. Three cards are picked at random.<\/p>\n

                                  \n
                                1. Suppose you know that the picked cards are [latex]Q[\/latex] of spades, [latex]K[\/latex] of hearts and [latex]Q[\/latex] of spades. Can you decide if the sampling was with or without replacement?<\/li>\n
                                2. Suppose you know that the picked cards are [latex]Q[\/latex] of spades, [latex]K[\/latex] of hearts, and [latex]J[\/latex] of spades. Can you decide if the sampling was with or without replacement?<\/li>\n<\/ol>\n
                                  Show Solution<\/span><\/p>\n
                                  \nSolution:<\/p>\n
                                    \n
                                  1. With replacement<\/li>\n
                                  2. No<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n

                                    <\/h3>\n
                                    \n

                                    Example<\/h3>\n

                                    You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of [latex]1, 2, 3, 4, 5, 6, 7, 8, 9, 10[\/latex],\u00a0[latex]J[\/latex] (jack), [latex]Q[\/latex] (queen), [latex]K[\/latex] (king) of that suit. [latex]S[\/latex] = spades, [latex]H[\/latex] = hearts, [latex]D[\/latex] = diamonds, [latex]C[\/latex] = clubs.<\/p>\n

                                      \n
                                    1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are [latex]QS[\/latex], [latex]1D[\/latex], [latex]1C[\/latex], [latex]QD[\/latex].<\/li>\n
                                    2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are [latex]KH[\/latex], [latex]7D[\/latex], [latex]6D[\/latex], [latex]KH[\/latex].<\/li>\n<\/ol>\n

                                      Which of [latex]1[\/latex] or [latex]2[\/latex] did you sample with replacement and which did you sample without replacement?<\/p>\n

                                      Show Solution<\/span><\/p>\n
                                      \nSolution:<\/p>\n
                                        \n
                                      1. Without replacement<\/li>\n
                                      2. With replacement<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n

                                         <\/p>\n

                                        \n

                                        This video provides a brief lesson on finding the probability of independent events.<\/p>\n