Compound Events 1
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Question 1 of 8
1. Question
A normal sixsided dice is rolled. Identify which of the following are mutually exclusive events:Hint
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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: `1,2,3,4,5,6`Even: `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: `1,2,3,4,5,6`Even: `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)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(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)$$ \mathsf{P(Red)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{15}}$$ Substitute values Next, find the probability of drawing a Green marblefavourable outcomes`=``4` (`4` Green)total outcomes`=``15` (`9` Red, `4` Green, `2` Black)$$ \mathsf{P(Green)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{15}}$$ Substitute values Finally, add the two probabilities$$ \mathsf{P(Red\:or\:Green)} $$ `=` $$\mathsf{P(Red)}+\mathsf{P(Green)}$$ Addition Rule `=` $$\frac{9}{15}+\frac{4}{15}$$ Substitute values `=` $$\frac{13}{15}$$ `(b)` Find the probability drawing a Red or Black marble.From part `(a)`, we have solved the probability of drawing a Red marble$$ \mathsf{P(Red)} $$ `=` $$\frac{9}{15}$$ Next, find the probability of drawing a Black marblefavourable outcomes`=``2` (`2` Black)total outcomes`=``15` (`9` Red, `4` Green, `2` Black)$$ \mathsf{P(Black)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{15}}$$ Substitute values Finally, add the two probabilities$$ \mathsf{P(Red\:or\:Black)} $$ `=` $$\mathsf{P(Red)}+\mathsf{P(Black)}$$ Addition Rule `=` $$\frac{9}{15}+\frac{2}{15}$$ Substitute values `=` $$\frac{11}{15}$$ `(a) 13/15``(b) 11/15` 

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)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(a)` Find the probability of drawing a Black `8` or Hearts.Start by finding the probability of drawing a Black `8`favourable outcomes`=``2` (`8` of Spades, `8` of Clubs)total outcomes`=``52` (a standard deck has `52` cards)$$ \mathsf{P(Black\:8)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{52}}$$ 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)$$ \mathsf{P(Hearts)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{13}}{\color{#007DDC}{52}}$$ Substitute values Finally, add the two probabilities$$ \mathsf{P(Black\:8\:or\:Hearts)} $$ `=` $$\mathsf{P(Black\:8)}+\mathsf{P(Hearts)}$$ Addition Rule `=` $$\frac{2}{52}+\frac{13}{52}$$ Substitute values `=` $$\frac{15}{52}$$ `(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)$$ \mathsf{P(King)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{52}}$$ 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)$$ \mathsf{P(Red)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{26}}{\color{#007DDC}{52}}$$ 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 `2/52` from the final probability.Finally, solve for the final probability$$ \mathsf{P(King\:or\:Red)} $$ `=` $$\mathsf{P(King)}+\mathsf{P(Red)}\frac{2}{52}$$ Addition Rule `=` $$\frac{4}{52}+\frac{26}{52}\frac{2}{52}$$ Substitute values `=` $$\frac{28}{52}$$ `=` $$\frac{7}{13}$$ `(a) 15/52``(b) 7/13` 

Question 4 of 8
4. Question
Find the probability of rolling `2` normal sixsided dice and getting:`(a)` Sum of `10``(b)` Sum of `7`Write fractions in the format “a/b”
`(a)` (1/12, 3/36)`(b)` (1/6, 6/36)
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(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 `10`This means there are `3` times that we can have a sum of `10`Finally, find the probability of getting a sum of `10`favourable outcomes`=``3`total outcomes`=``36`$$ \mathsf{P(sum}\:10) $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{12}$$ `(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 `7`This means there are `6` times that we can have a sum of `7`Finally, find the probability of getting a sum of `7`favourable outcomes`=``6`total outcomes`=``36`$$ \mathsf{P(sum}\:7) $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{6}$$ `(a) 1/12``(b) 1/6` 

Question 5 of 8
5. Question
A normal sixsided 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
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Set 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`=``3`total outcomes`=``12`$$ \mathsf{P(Even\:and\:Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$ Substitute values `=` $$\frac{1}{4}$$ `1/4` 
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
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(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`=``1`total outcomes`=``4`$$ \mathsf{P(2\:Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ 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`=``2`total outcomes`=``4`$$ \mathsf{P(Heads\:and\:Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{4}}$$ Substitute values `=` $$\frac{1}{2}$$ `(a) 1/4``(b) 1/2` 

Question 7 of 8
7. Question
Find the probability of rolling `2` normal sixsided dice and getting:`(a)` Sum of `9` or `10``(b)` Sum of `6` or `12`Write fractions in the format “a/b”
`(a)` (7/36)`(b)` (1/6, 6/36)
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Addition Rule (Mutually Exclusive Events)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(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 `10`Having 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=7`total outcomes`=``36`$$ \mathsf{P(sum\:of\:9\:or\:10)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{7}}{\color{#007DDC}{36}}$$ 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 `12`Having 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=6`total outcomes`=``36`$$ \mathsf{P(sum\:of\:6\:or\:12)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{6}$$ `(a) 7/36``(b) 1/6` 

Question 8 of 8
8. Question
Find the probability of rolling `2` normal sixsided dice and getting `1` or `5` dots on the diceWrite fractions in the format “a/b” (5/9, 20/36)
Hint
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Addition Rule (Mutually Exclusive Events)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Set 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 appear`1` dot appear `11` times`5` 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 `2/36` from the final probability.Solve for the final probability using Addition Rulefavourable outcomes`=``11+11=22`total outcomes`=``36`$$ \mathsf{P(1\:or\:5)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}\frac{2}{36}$$ Probability Formula `=` $$\frac{\color{#e65021}{22}}{\color{#007DDC}{36}}\frac{2}{36}$$ Substitute values `=` $$\frac{20}{36}$$ `=` $$\frac{5}{9}$$ `5/9`