Compound Events (Addition Rule) 1
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Question 1 of 8
1. Question
A normal six-sided dice is rolled. Identify which of the following are mutually exclusive events:- 1.
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2.
Even or Odd
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Events are mutually exclusive if they cannot occur simultaneously.Rolling the dice and getting an Even or Odd number.Compare the results for each separate eventSample Space of rolling a dice: 1,2,3,4,5,6Even: 2,4,6 Odd: 1,3,5 None of the results occurred simultaneously or is repeated. Hence, these are mutually exclusive events.Rolling the dice and getting an Even or >5 number of dots.Compare the results for each separate eventSample Space of rolling a dice: 1,2,3,4,5,6Even: 2,4,6 >5: 6 6 dots occurred simultaneously and is repeated on the two events. Hence, these are NOT mutually exclusive events.Getting an Even or Odd number of dots is a mutually exclusive event -
Question 2 of 8
2. Question
A jar contains 9 Red, 4 Green and 2 Black marbles. Find the probability of drawing a marble at random and getting:(a) Red or Green(b) Red or BlackWrite fractions in the format “a/b”-
(a) (13/15)(b) (11/15)
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Addition Rule (Mutually Exclusive Events)
P(AorB)=P(A)+P(B)Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
P(E)=favourableoutcomestotaloutcomes(a) Find the probability drawing a Red or Green marble.Start by finding the probability of drawing a Red marblefavourable outcomes=9 (9 Red)total outcomes=15 (9 Red, 4 Green, 2 Black)P(Red) = favourableoutcomestotaloutcomes Probability Formula = 915 Substitute values Next, find the probability of drawing a Green marblefavourable outcomes=4 (4 Green)total outcomes=15 (9 Red, 4 Green, 2 Black)P(Green) = favourableoutcomestotaloutcomes Probability Formula = 415 Substitute values Finally, add the two probabilitiesP(RedorGreen) = P(Red)+P(Green) Addition Rule = 915+415 Substitute values = 1315 (b) Find the probability drawing a Red or Black marble.From part (a), we have solved the probability of drawing a Red marbleP(Red) = 915 Next, find the probability of drawing a Black marblefavourable outcomes=2 (2 Black)total outcomes=15 (9 Red, 4 Green, 2 Black)P(Black) = favourableoutcomestotaloutcomes Probability Formula = 215 Substitute values Finally, add the two probabilitiesP(RedorBlack) = P(Red)+P(Black) Addition Rule = 915+215 Substitute values = 1115 (a) 1315(b) 1115 -
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Question 3 of 8
3. Question
Find the probability of drawing from a standard deck of cards and getting:(a) Black 8 or Hearts(b) King or RedWrite fractions in the format “a/b”-
(a) (15/52)(b) (7/13, 28/52)
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Addition Rule (Mutually Exclusive Events)
P(AorB)=P(A)+P(B)Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
P(E)=favourableoutcomestotaloutcomes(a) Find the probability of drawing a Black 8 or Hearts.Start by finding the probability of drawing a Black 8favourable outcomes=2 (8 of Spades, 8 of Clubs)total outcomes=52 (a standard deck has 52 cards)P(Black8) = favourableoutcomestotaloutcomes Probability Formula = 252 Substitute values Next, find the probability of drawing a Hearts cardfavourable outcomes=13 (13 Hearts cards)total outcomes=52 (a standard deck has 52 cards)P(Hearts) = favourableoutcomestotaloutcomes Probability Formula = 1352 Substitute values Finally, add the two probabilitiesP(Black8orHearts) = P(Black8)+P(Hearts) Addition Rule = 252+1352 Substitute values = 1552 (b) Find the probability of drawing a King or Red card.Start by finding the probability of drawing a Kingfavourable outcomes=4 (each of the suits have a King card)total outcomes=52 (a standard deck has 52 cards)P(King) = favourableoutcomestotaloutcomes Probability Formula = 452 Substitute values Next, find the probability of drawing a Red cardfavourable outcomes=26 (13 Hearts, 13 Diamonds)total outcomes=52 (a standard deck has 52 cards)P(Red) = favourableoutcomestotaloutcomes Probability Formula = 2652 Substitute values Remember that the events need to be mutually exclusive which means the same card can’t be counted twice.2 of the cards are counted twice, which are King of Hearts and King of DiamondsHence, subtract 252 from the final probability.Finally, solve for the final probabilityP(KingorRed) = P(King)+P(Red)−252 Addition Rule = 452+2652−252 Substitute values = 2852 = 713 (a) 1552(b) 713 -
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Question 4 of 8
4. Question
Find the probability of rolling 2 normal six-sided dice and getting:(a) Sum of 10(b) Sum of 7Write fractions in the format “a/b”-
(a) (1/12, 3/36)(b) (1/6, 6/36)
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Probability Formula
P(E)=favourableoutcomestotaloutcomes(a) Find the probability of getting a Sum of 10.Set up a lattice showing all possible sums for the two diceThis means there are 36 possible outcomesNow, count how many times we can have a sum of 10This means there are 3 times that we can have a sum of 10Finally, find the probability of getting a sum of 10favourable outcomes=3total outcomes=36P(sum10) = favourableoutcomestotaloutcomes Probability Formula = 336 Substitute values = 112 (b) Find the probability of getting a Sum of 7.Use the lattice from part (a) and count how many times we can have a sum of 7This means there are 6 times that we can have a sum of 7Finally, find the probability of getting a sum of 7favourable outcomes=6total outcomes=36P(sum7) = favourableoutcomestotaloutcomes Probability Formula = 636 Substitute values = 16 (a) 112(b) 16 -
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Question 5 of 8
5. Question
A normal six-sided dice is rolled and a fair coin is tossed. Find the probability of getting an Even number of dots and Heads.Write fractions in the format “a/b”- (¼, 1/4, 3/12)
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Probability Formula
P(E)=favourableoutcomestotaloutcomesSet up a lattice showing all possible results for rolling a dice and tossing a coinThis means there are 12 possible outcomesNow, count how many times we can have an Even number of dots and HeadsThis means there are 3 times that we can have an Even number of dots and HeadsFinally, solve for the probabilityfavourable outcomes=3total outcomes=12P(EvenandHeads) = favourableoutcomestotaloutcomes Probability Formula = 312 Substitute values = 14 14 -
Question 6 of 8
6. Question
Find the probability of tossing 2 coins and getting:(a) 2 Tails(b) Heads and TailsWrite fractions in the format “a/b”-
(a) (1/4, ¼)(b) (½, 1/2, 2/4)
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Probability Formula
P(E)=favourableoutcomestotaloutcomes(a) Find the probability of getting 2 Tails.Set up a lattice showing all possible results for tossing 2 coinsThis means there are 4 possible outcomesNow, count how many times we can have 2 TailsThis means there is 1 time that we can have 2 TailsFinally, find the probability of getting 2 Tailsfavourable outcomes=1total outcomes=4P(2Tails) = favourableoutcomestotaloutcomes Probability Formula = 14 Substitute values (b) Find the probability of getting Heads and Tails.Use the lattice from part (a) and count how many times we can have Heads and TailsThis means there are 2 times that we can have Heads and TailsFinally, find the probability of getting Heads and Tailsfavourable outcomes=2total outcomes=4P(HeadsandTails) = favourableoutcomestotaloutcomes Probability Formula = 24 Substitute values = 12 (a) 14(b) 12 -
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Question 7 of 8
7. Question
Find the probability of rolling 2 normal six-sided dice and getting:(a) Sum of 9 or 10(b) Sum of 6 or 12Write fractions in the format “a/b”-
(a) (7/36)(b) (1/6, 6/36)
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Addition Rule (Mutually Exclusive Events)
P(AorB)=P(A)+P(B)Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
P(E)=favourableoutcomestotaloutcomes(a) Find the probability of getting a Sum of 9 or 10.Set up a lattice showing all possible sums for the two diceThis means there are 36 possible outcomesNow, count how many times we can have a sum of 9 or 10Having a sum of 9 appears 4 timeHaving a sum of 10 appears 3 timesFinally, find the probability of getting a sum of 9 or 10 using Addition Rulefavourable outcomes=4+3=7total outcomes=36P(sumof9or10) = favourableoutcomestotaloutcomes Probability Formula = 736 Substitute values (b) Find the probability of getting a Sum of 6 or 12.Use the lattice from part (a) and count how many times we can have a sum of 6 or 12Having a sum of 6 appears 5 timesHaving a sum of 12 appears 1 timesFinally, find the probability of getting a sum of 6 or 12 using Addition Rulefavourable outcomes=5+1=6total outcomes=36P(sumof6or12) = favourableoutcomestotaloutcomes Probability Formula = 636 Substitute values = 16 (a) 736(b) 16 -
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Question 8 of 8
8. Question
Find the probability of rolling 2 normal six-sided dice and getting 1 or 5 dots on the diceWrite fractions in the format “a/b”- (5/9, 20/36)
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Addition Rule (Mutually Exclusive Events)
P(AorB)=P(A)+P(B)Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
P(E)=favourableoutcomestotaloutcomesSet up a lattice showing all possible results for rolling two diceThis means there are 36 possible outcomesNow, count how many times 1 or 5 dots will appear1 dot appear 11 times5 dots appear 11 timesRemember that each roll must only be counted once.2 rolls are counted twice since they result to having both 1 AND 5 dots.Hence, subtract 236 from the final probability.Solve for the final probability using Addition Rulefavourable outcomes=11+11=22total outcomes=36P(1or5) = favourableoutcomestotaloutcomes−236 Probability Formula = 2236−236 Substitute values = 2036 = 59 59
Quizzes
- Simple Probability (Theoretical) 1
- Simple Probability (Theoretical) 2
- Simple Probability (Theoretical) 3
- Simple Probability (Theoretical) 4
- Complementary Probability
- Compound Events (Addition Rule) 1
- Compound Events (Addition Rule) 2
- Venn Diagrams (Mutually Inclusive)
- Independent Events 1
- Independent Events 2
- Dependent Events (Conditional Probability)
- Probability Tree (Independent Events) 1
- Probability Tree (Independent Events) 2
- Probability Tree (Dependent Events)