mutually exclusive and independent events formula

To be mutually exclusive, \(P(\text{C AND E})\) must be zero.Let event \(\text{A} =\) learning Spanish. (Hint: What is \(P(\text{A AND B})\)? \(P(\text{E}) = \dfrac{2}{4}\).Let \(\text{F} =\) the event of getting at most one tail (zero or one tail).Let \(\text{G} =\) the event of getting two faces that are the same.Let \(\text{H} =\) the event of getting a head on the first flip followed by a head or tail on the second flip.Are \(\text{F}\) and \(\text{G}\) mutually exclusive?Let \(\text{J} =\) the event of getting all tails. Show \(P(\text{G AND H}) = P(\text{G})P(\text{H})\).\(P(\text{G|H}) = \dfrac{P(\text{G AND H})}{P(\text{H})} = \dfrac{0.3}{0.5} = 0.6 = P(\text{G})\),\(P(\text{G})P(\text{H}) = (0.6)(0.5) = 0.3 = P(\text{G AND H})\).The sample space is \(\text{S} = \{R1, R2, R3, R4, R5, R6, G1, G2, G3, G4\}\).Are \(\text{C}\) and \(\text{D}\) independent?Are \(\text{C}\) and \(\text{D}\) mutually exclusive?Yes, because \(P(\text{C|D}) = P(\text{C})\).No, because \(P(\text{C AND D})\) is not equal to zero.\(P(\text{D|C}) = \dfrac{P(\text{C AND D})}{P(\text{C})} = \dfrac{0.225}{0.75} = 0.3\).Are \(\text{B}\) and \(\text{D}\) independent?Are \(\text{B}\) and \(\text{D}\) mutually exclusive?\(\text{E} =\) even-numbered card is drawn.\(P(\text{R}) = \dfrac{3}{8}\).

It consists of four suits. Content produced by.\(\text{E}\) and \(\text{F}\) are mutually exclusive events. \(\text{A}\) and \(\text{C}\) do not have any numbers in common so \(P(\text{A AND C}) = 0\). Fifty percent of all students in the class have long hair. Also, \(P(\text{A}) = \dfrac{3}{6}\) and \(P(\text{B}) = \dfrac{3}{6}\).Let event \(\text{C} =\) odd faces larger than two. If two events are NOT independent, then we say that they are dependent.Sampling may be done with replacement or without replacement (Figure \(\PageIndex{1}\)):If it is not known whether \(\text{A}\) and \(\text{B}\) are independent or dependent, assume they are dependent until you can show otherwise.You have a fair, well-shuffled deck of 52 cards. Are they mutually exclusive?So \(P(\text{B})\) does not equal \(P(\text{B|A})\) which means that \(\text{B} and \text{A}\) are not independent (wearing blue and rooting for the away team are not independent). You reach into the box (you cannot see into it) and draw one card.The sample space \(S = R1, R2, R3, B1, B2, B3, B4, B5\).Let \(\text{A}\) be the event that a fan is rooting for the away team.Let \(\text{B}\) be the event that a fan is wearing blue.Are the events of rooting for the away team and wearing blue independent? Let \(\text{F}\) be the event that a student is female. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), \(\text{K}\) (king) of that suit.Suppose you pick three cards with replacement. (There are three even-numbered cards, \(R2, B2\), and \(B4\). Then \(\text{A} = \{1, 3, 5\}\). (You cannot draw one card that is both red and blue. Some of the following questions do not have enough information for you to answer them. Answer the same question for sampling with replacement.a. Because you have picked the cards without replacement, you cannot pick the same card twice.You have a fair, well-shuffled deck of 52 cards. \(\text{B}\) and \(\text{C}\) have no members in common because you cannot have all tails and all heads at the same time. Adopted or used LibreTexts for your course? Your cards are \(\text{QS}, 1\text{D}, 1\text{C}, \text{QD}\).Suppose you pick four cards and put each card back before you pick the next card. (There are five blue cards: \(B1, B2, B3, B4\), and \(B5\). \(\text{J}\) and \(\text{H}\) are mutually exclusive.A box has two balls, one white and one red. Which of the following outcomes are possible? The third card is the \(\text{J}\) of spades. Find the probability of the following events:Roll one fair, six-sided die. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), and \(\text{K}\) (king) of that suit. ),\(P(\text{E|B}) = \dfrac{2}{5}\). \(\text{F}\) and \(\text{G}\) are not mutually exclusive.Getting all tails occurs when tails shows up on both coins (\(TT\)). Let event \(\text{A} =\) a face is odd. The sample space is {1, 2, 3, 4, 5, 6}. Show that \(P(\text{G|H}) = P(\text{G})\).b. Mutually exclusive events do not overlap at any point, therefore $P(A \cap B)=0$.

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