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Probability: Mutually Exclusive Events 1. There are 3 red, 4 black and 5 blue cubes in a bag. A cube is selected at random. What is the probability of selecting a red or a blue cube? Since there are 8 cubes that are red or blue out of a total of 12, the answer must be 8/12. P(red or blue) = 8/12 Note that we could have written down the individual probabilities and added them to get the same answer like so: P(red or blue) = P(red) + P(blue) = 3/12 + 5/12 = 8/12 2. Find the probability of selecting an ace or a king from a pack of 52 cards. Simply by counting the outcomes needed (4 aces and 4 kings), the answer is 8/52. P(ace or king) = 8/52. Or, as before, we get the same answer by adding the individual probabilities. P(ace or king) = P(ace) + P(king) = 4 /52 + 4/52 = 8/52

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In these two examples, the events that occurred could not have happened at the same time. In 1. A red and a blue die could not have been chosen at the same time. In 2. An ace and a king could not have been chosen at the same time. Mutually Exclusive Events Events that cannot happen at the same time are called MUTUALLY EXCLUSIVE. The OR LAW When events are mutually exclusive we can always add the probabilities. For mutually exclusive events: P(A or B) = P(A) + P(B)

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The OR LAW For mutually exclusive events: P(A or B) = P(A) + P(B) 0 Impossible 1 Certain ½ When using the Or Law and adding the individual probabilities, the cumulative effect increases the likelihood of the combined events happening. P(red) = 3/12 P(red or blue ) = 3/12 + 5/12 = 8/12 P(red or blue or black ) = 3/12 + 5/12 + 4/12 = 1 This may seem obvious but it is worth stating explicitly since there is another law to come later, (the AND LAW) where the probabilities are reduced and sometimes confusion occurs between the two. 123

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If the events in question are not mutually exclusive then the OR LAW cannot be used and the probabilities cannot be added. Events that are NOT Mutually Exclusive Consider the problem of taking a card at random from a pack. What is the probability of selecting a diamond or a queen? These events are not mutually exclusive because the queen of diamonds is both a queen and a diamond.

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Events that are NOT Mutually Exclusive If we tried to use the OR LAW then: P(diamond or queen) = P(diamond) + P(queen) = 13/52 + 4/52 = 17/52 Whereas there are in fact only 16 cards that are diamonds or queens. Therefore P(diamond or queen) = 16/52

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Consider the problem of spinning the pointer on the spinner. What is the probability of the pointer indicating a green colour or a number 5. These events are not mutually exclusive since the pointer can indicate green and 5 at the same time. If we tried to use the OR LAW then: P(Green or 5) = P(Green) + P(5) = 1/6 + 2/6 = 3/6 Whereas there are only 2 areas of the spinner that show green or a 5 Therefore P(Green or 5) = 2/6

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A single card is chosen at random from the full deck of cards. Work out the probabilities of the following events and simplify your answer where necessary. (a) P(club) (b) P(picture card) (c) P(queen) (d) P(heart or spade) (e) P(9 or 2) (a) P(club) = 13/52 = 1/4 (b) P(picture card) = 12/52 = 3/13 (c) P(queen) = 4/52 = 1/13 (d) P(heart or spade) = 13/ /52 = 26/52 = 1/2 (e) P(9 or 2) = 4/52 + 4/52 = 8/52 = 2/13

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(a) P(3 or 7 or king) (b) P(heart or diamond or spade) (c) P(diamond or ace) (a) P(3 or 7 or king) = 4/52 + 4/52 + 4/52 = 12/52 = 3/13 (b) P(heart or diamond or spade) = 13/ / /52 = 39/52 = 3/4 (c) P(diamond or ace) = 16/52 = 4/13 (events are not mutually exclusive)

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A single card is chosen at random from the depleted deck of cards. Work out the probabilities of the following events and simplify your answer where necessary. (a) P(spade) (b) P(picture card) (c) P(king) (d) P(heart or spade) (e) P(4 or 7) (a) P(spade) = 10/40 = 1/4 (b) P(picture card) = 11/40 (c) P(king) = 4/40 =1/10 (d) P(heart or spade) = 11/ /40 = 21/40 (e) P(4 or 7) = 1/40 + 4/40 = 5/40 = 1/8

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(a) P(2 or 7 or king) (b) P(heart or diamond or spade) (c) P(diamond or jack) (a) P(2 or 7 or king) = 3/48 + 4/48 + 2/48 = 9/48 = 3/16 (b) P(heart or diamond or spade) = 12/ / /48 = 37/48 (c) P(diamond or jack) = 15/48 = 5/16 (events are not mutually exclusive)

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Player 1 lays a card at random from his hand. What are the probabilities of the card being: (a) A king or a 7. (b) A heart or a club. (c) A heart or king. Player 1 (a) P(king or 7) = 2/7 + 1/7 = 3/7(b) P(heart or club) = 3/7 + 2/7 = 5/7 (c) P(heart or king) = 4/7 (events not mutually exclusive) Player 2

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Player 2 lays a card at random from his hand. What are the probabilities of the card being: (a) A queen or a 7. (b) A diamond or a club (c) A club or a 7 (a) P(queen or 7) = 2/8 + 2/8 = 4/8 (1/2)(b) P(diamond or club) = 2/8 + 3/8 = 5/8 (c) P(club or 7) = 4/8 (1/2) (events not mutually exclusive) Player 1 Player 2

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Probability: Mutually Exclusive Events 1. There are 3 red, 4 black and 5 blue cubes in a bag. A cube is selected at random. What is the probability of selecting a red or a blue cube? Since there are 8 cubes that are red or blue out of a total of 12, the answer must be 8/12. Note that we could have written down the individual probabilities and added them to get the same answer like so: 2. Find the probability of selecting an ace or a king from a pack of 52 cards. Simply by counting the outcomes needed (4 aces and 4 kings), the answer is 8/52. Or, as before, we get the same answer by adding the individual probabilities.

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In these two examples, the events that occurred could In 1. A red and a blue die could not have been chosen at the same time. In 2. An ace and a king could not have been chosen at the same time. Mutually Exclusive Events Events that cannot happen at the same time are called. The OR LAW When events are mutually exclusive we can always add the probabilities. For mutually exclusive events:

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If the events in question are not mutually exclusive then the OR LAW cannot be used and the probabilities cannot be added. Events that are NOT Mutually Exclusive Consider the problem of taking a card at random from a pack. What is the probability of selecting a diamond or a queen?

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Events that are NOT Mutually Exclusive If we tried to use the OR LAW then: P(diamond or queen) = P(diamond) + P(queen) = Whereas there are in fact only cards that are diamonds or queens. Therefore P(diamond or queen) =

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Consider the problem of spinning the pointer on the spinner. What is the probability of the pointer indicating a green colour or a number 5. These events are not mutually exclusive since the pointer can indicate green and 5 at the same time. If we tried to use the OR LAW then: Whereas there are only 2 areas of the spinner that show green or a 5 Therefore P(Green or 5) =

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Player 2 lays a card at random from his hand. What are the probabilities of the card being: (a) A queen or a 7. (b) A diamond or a club (c) A club or a 7 (a) P(queen or 7) =(b) P(diamond or club) = (c) P(club or 7) = Player 1 Player 2 Homework: page 312 – 313 # 1, 4, 6, 7, 9, 11, 12, Page 340 – 343 # 1, 3, 4, 5, 7, 8, 13

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Using the roulette wheel, find the probabilities of the ball landing on: (a) Number 7 (b) An odd number (c) A prime Number (d) The colour red (e) A number ending in 4 or 9 (f) A red or 15 (g) A black or 10 (a) P(7) = 1/37(b) P(odd) = 18/37 (c) P(prime) = 11/37 (d) P(red) = 18/37 (e) P(number ending in 4 or 9) = 4/37 + 3/37 = 7/37 (f) P(red or 15) = 18/37 + 1/37 = 19/37 (g) P(black or 10) = 18/37 events are not mutually exclusive

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